Showing a conditional statement is a tautology without using a truth table.












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I wanted to show that [(p→q)∧(q→r)]→(p→r) is a tautology without using a truth table. This is what I got so far:



[(p→q)∧(q→r)] → (p→r)



=> ¬[(¬p v q) ∧ (¬q v r)] v (¬pvr) (logical equivalence)



=> [¬(¬p v q) v ¬(¬qvr)] v (¬pvr) (demorgan's law)



=> [(p ∧ ¬q) v (q∧¬r)] v (¬pvr) (demogran's law)



I can't seem to figure out what comes after this step. Can someone help me?










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    2












    $begingroup$


    I wanted to show that [(p→q)∧(q→r)]→(p→r) is a tautology without using a truth table. This is what I got so far:



    [(p→q)∧(q→r)] → (p→r)



    => ¬[(¬p v q) ∧ (¬q v r)] v (¬pvr) (logical equivalence)



    => [¬(¬p v q) v ¬(¬qvr)] v (¬pvr) (demorgan's law)



    => [(p ∧ ¬q) v (q∧¬r)] v (¬pvr) (demogran's law)



    I can't seem to figure out what comes after this step. Can someone help me?










    share|cite|improve this question







    New contributor




    Nev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      2












      2








      2


      1



      $begingroup$


      I wanted to show that [(p→q)∧(q→r)]→(p→r) is a tautology without using a truth table. This is what I got so far:



      [(p→q)∧(q→r)] → (p→r)



      => ¬[(¬p v q) ∧ (¬q v r)] v (¬pvr) (logical equivalence)



      => [¬(¬p v q) v ¬(¬qvr)] v (¬pvr) (demorgan's law)



      => [(p ∧ ¬q) v (q∧¬r)] v (¬pvr) (demogran's law)



      I can't seem to figure out what comes after this step. Can someone help me?










      share|cite|improve this question







      New contributor




      Nev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I wanted to show that [(p→q)∧(q→r)]→(p→r) is a tautology without using a truth table. This is what I got so far:



      [(p→q)∧(q→r)] → (p→r)



      => ¬[(¬p v q) ∧ (¬q v r)] v (¬pvr) (logical equivalence)



      => [¬(¬p v q) v ¬(¬qvr)] v (¬pvr) (demorgan's law)



      => [(p ∧ ¬q) v (q∧¬r)] v (¬pvr) (demogran's law)



      I can't seem to figure out what comes after this step. Can someone help me?







      discrete-mathematics proof-verification logic






      share|cite|improve this question







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      Nev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question







      New contributor




      Nev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




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      asked 4 hours ago









      NevNev

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      183




      New contributor




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      New contributor





      Nev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          4 Answers
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          1












          $begingroup$

          Notice that



          $ [(p land neg q) lor (qland neg r)] lor (neg plor r)$



          is one big disjunction, so you can drop parentheses:



          $ (p land neg q) lor (qland neg r) lor neg plor r$



          Now, if you have:



          Reduction



          $p lor (neg p land q) equiv p lor q$



          then you can apply that:



          $ (p land neg q) lor (qland neg r) lor neg plor r equiv$



          $neg q lor q lor neg p lor r equiv$



          $top lor neg p lor r equiv$



          $top$



          But if you don't have Reduction:



          $ (p land neg q) lor (qland neg r) lor neg p lor r equiv$



          $((p lor neg p) land (neg q lor neg p)) lor ((q lor r) land (neg r lor r)) equiv$



          $(top land (neg q lor neg p)) lor ((q lor r) land top) equiv$



          $neg q lor neg p lor q lor r$



          $top lor neg p lor r equiv$



          $top$






          share|cite|improve this answer









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          • $begingroup$
            Thank you so much, makes a lot of sense now
            $endgroup$
            – Nev
            2 hours ago










          • $begingroup$
            @nev you're welcome!
            $endgroup$
            – Bram28
            1 hour ago



















          1












          $begingroup$

          Expand the expression $(p wedge neg q) vee (q wedge neg r)$ by distributing the $vee$ over the $wedge$:
          begin{align*}
          &(p wedge neg q) vee (q wedge neg r)\
          &[(p wedge neg q) vee q] wedge [(p wedge neg q) vee neg r]\
          &[(p vee q ) wedge (neg q vee q)] wedge [(p vee neg r ) wedge (neg q vee neg r)]\
          &[(p vee q) wedge top ] wedge[(p vee neg r) wedge (neg q vee neg r)]\
          &(p vee q) wedge [(p vee neg r) wedge (neg q vee neg r)].
          end{align*}

          Now overall we have ${(p vee q) wedge [(p vee neg r) wedge (neg q vee neg r)]}vee (neg p vee r)$. If we distribute the $vee$ over the $wedge$, we get $[(p vee q) vee ( neg p vee r)]wedge{[(p vee neg r) wedge (neg q vee neg r)]vee(neg p vee r)}$. Focusing on the first half, you can manipulate $(p vee q) vee ( neg p vee r)$ to get $top$ by shuffling parentheses around to get $(p vee neg p) vee (q vee r)$ (I'll leave that to you).



          So we are left with $[(p vee neg r) wedge (neg q vee neg r)] vee (neg p vee r)$. Again let's distribute the $vee$ over the $wedge$:
          begin{align*}
          &[(p vee neg r) vee (neg p vee r)] wedge [(neg q vee neg r) vee (neg p vee r)].
          end{align*}

          Again both halves of this can be manipulated to get $top$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$


            I can't seem to figure out what comes after this step. Can someone help me?




            Yes.



            $$begin{align}vdotsquad\iff&~big((p land lnot q) lor (qlandlnot r)big) lor (lnot plor r)
            \[1ex]iff &~big(lnot plor (pland lnot q)big)lor big(rlor (lnot rland q)big)&quadtextsf{(Commutation and Association)}\vdotsquadend{align}$$






            share|cite|improve this answer









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              1












              $begingroup$

              You can use Double Distribution to get
              $$[(p lor q)land(q lor lnot q)land(lnot q lor lnot r)land(p lor lnot r)]lor (lnot p lor r)$$
              $q lor lnot q$ is a Tautology so this becomes
              $$[(p lor q)land(lnot q lor lnot r)land(p lor lnot r)]lor (lnot p lor r)$$
              which by distribution is
              $$[(p lor q)land[lnot rland(p lor lnot q)]]lor (lnot p lor r)$$
              Association gives
              $$[lnot r land [(p lor q)land(p lor lnot q)]]lor (lnot p lor r)$$
              Distribution again gives
              $$[lnot r land [p lor(q land lnot q)]]lor (lnot p lor r)$$
              $q land lnot q$ is a contradiction so this becomes
              $$(lnot r land p)lor (lnot p lor r)$$
              Which by DeMorgan's Law is
              $$(lnot r land p)lor lnot(lnot r land p)$$
              Which is a tautology.






              share|cite|improve this answer









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              • $begingroup$
                Thanks for the help!!
                $endgroup$
                – Nev
                2 hours ago











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              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

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              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Notice that



              $ [(p land neg q) lor (qland neg r)] lor (neg plor r)$



              is one big disjunction, so you can drop parentheses:



              $ (p land neg q) lor (qland neg r) lor neg plor r$



              Now, if you have:



              Reduction



              $p lor (neg p land q) equiv p lor q$



              then you can apply that:



              $ (p land neg q) lor (qland neg r) lor neg plor r equiv$



              $neg q lor q lor neg p lor r equiv$



              $top lor neg p lor r equiv$



              $top$



              But if you don't have Reduction:



              $ (p land neg q) lor (qland neg r) lor neg p lor r equiv$



              $((p lor neg p) land (neg q lor neg p)) lor ((q lor r) land (neg r lor r)) equiv$



              $(top land (neg q lor neg p)) lor ((q lor r) land top) equiv$



              $neg q lor neg p lor q lor r$



              $top lor neg p lor r equiv$



              $top$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Thank you so much, makes a lot of sense now
                $endgroup$
                – Nev
                2 hours ago










              • $begingroup$
                @nev you're welcome!
                $endgroup$
                – Bram28
                1 hour ago
















              1












              $begingroup$

              Notice that



              $ [(p land neg q) lor (qland neg r)] lor (neg plor r)$



              is one big disjunction, so you can drop parentheses:



              $ (p land neg q) lor (qland neg r) lor neg plor r$



              Now, if you have:



              Reduction



              $p lor (neg p land q) equiv p lor q$



              then you can apply that:



              $ (p land neg q) lor (qland neg r) lor neg plor r equiv$



              $neg q lor q lor neg p lor r equiv$



              $top lor neg p lor r equiv$



              $top$



              But if you don't have Reduction:



              $ (p land neg q) lor (qland neg r) lor neg p lor r equiv$



              $((p lor neg p) land (neg q lor neg p)) lor ((q lor r) land (neg r lor r)) equiv$



              $(top land (neg q lor neg p)) lor ((q lor r) land top) equiv$



              $neg q lor neg p lor q lor r$



              $top lor neg p lor r equiv$



              $top$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Thank you so much, makes a lot of sense now
                $endgroup$
                – Nev
                2 hours ago










              • $begingroup$
                @nev you're welcome!
                $endgroup$
                – Bram28
                1 hour ago














              1












              1








              1





              $begingroup$

              Notice that



              $ [(p land neg q) lor (qland neg r)] lor (neg plor r)$



              is one big disjunction, so you can drop parentheses:



              $ (p land neg q) lor (qland neg r) lor neg plor r$



              Now, if you have:



              Reduction



              $p lor (neg p land q) equiv p lor q$



              then you can apply that:



              $ (p land neg q) lor (qland neg r) lor neg plor r equiv$



              $neg q lor q lor neg p lor r equiv$



              $top lor neg p lor r equiv$



              $top$



              But if you don't have Reduction:



              $ (p land neg q) lor (qland neg r) lor neg p lor r equiv$



              $((p lor neg p) land (neg q lor neg p)) lor ((q lor r) land (neg r lor r)) equiv$



              $(top land (neg q lor neg p)) lor ((q lor r) land top) equiv$



              $neg q lor neg p lor q lor r$



              $top lor neg p lor r equiv$



              $top$






              share|cite|improve this answer









              $endgroup$



              Notice that



              $ [(p land neg q) lor (qland neg r)] lor (neg plor r)$



              is one big disjunction, so you can drop parentheses:



              $ (p land neg q) lor (qland neg r) lor neg plor r$



              Now, if you have:



              Reduction



              $p lor (neg p land q) equiv p lor q$



              then you can apply that:



              $ (p land neg q) lor (qland neg r) lor neg plor r equiv$



              $neg q lor q lor neg p lor r equiv$



              $top lor neg p lor r equiv$



              $top$



              But if you don't have Reduction:



              $ (p land neg q) lor (qland neg r) lor neg p lor r equiv$



              $((p lor neg p) land (neg q lor neg p)) lor ((q lor r) land (neg r lor r)) equiv$



              $(top land (neg q lor neg p)) lor ((q lor r) land top) equiv$



              $neg q lor neg p lor q lor r$



              $top lor neg p lor r equiv$



              $top$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 2 hours ago









              Bram28Bram28

              60.8k44590




              60.8k44590












              • $begingroup$
                Thank you so much, makes a lot of sense now
                $endgroup$
                – Nev
                2 hours ago










              • $begingroup$
                @nev you're welcome!
                $endgroup$
                – Bram28
                1 hour ago


















              • $begingroup$
                Thank you so much, makes a lot of sense now
                $endgroup$
                – Nev
                2 hours ago










              • $begingroup$
                @nev you're welcome!
                $endgroup$
                – Bram28
                1 hour ago
















              $begingroup$
              Thank you so much, makes a lot of sense now
              $endgroup$
              – Nev
              2 hours ago




              $begingroup$
              Thank you so much, makes a lot of sense now
              $endgroup$
              – Nev
              2 hours ago












              $begingroup$
              @nev you're welcome!
              $endgroup$
              – Bram28
              1 hour ago




              $begingroup$
              @nev you're welcome!
              $endgroup$
              – Bram28
              1 hour ago











              1












              $begingroup$

              Expand the expression $(p wedge neg q) vee (q wedge neg r)$ by distributing the $vee$ over the $wedge$:
              begin{align*}
              &(p wedge neg q) vee (q wedge neg r)\
              &[(p wedge neg q) vee q] wedge [(p wedge neg q) vee neg r]\
              &[(p vee q ) wedge (neg q vee q)] wedge [(p vee neg r ) wedge (neg q vee neg r)]\
              &[(p vee q) wedge top ] wedge[(p vee neg r) wedge (neg q vee neg r)]\
              &(p vee q) wedge [(p vee neg r) wedge (neg q vee neg r)].
              end{align*}

              Now overall we have ${(p vee q) wedge [(p vee neg r) wedge (neg q vee neg r)]}vee (neg p vee r)$. If we distribute the $vee$ over the $wedge$, we get $[(p vee q) vee ( neg p vee r)]wedge{[(p vee neg r) wedge (neg q vee neg r)]vee(neg p vee r)}$. Focusing on the first half, you can manipulate $(p vee q) vee ( neg p vee r)$ to get $top$ by shuffling parentheses around to get $(p vee neg p) vee (q vee r)$ (I'll leave that to you).



              So we are left with $[(p vee neg r) wedge (neg q vee neg r)] vee (neg p vee r)$. Again let's distribute the $vee$ over the $wedge$:
              begin{align*}
              &[(p vee neg r) vee (neg p vee r)] wedge [(neg q vee neg r) vee (neg p vee r)].
              end{align*}

              Again both halves of this can be manipulated to get $top$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Expand the expression $(p wedge neg q) vee (q wedge neg r)$ by distributing the $vee$ over the $wedge$:
                begin{align*}
                &(p wedge neg q) vee (q wedge neg r)\
                &[(p wedge neg q) vee q] wedge [(p wedge neg q) vee neg r]\
                &[(p vee q ) wedge (neg q vee q)] wedge [(p vee neg r ) wedge (neg q vee neg r)]\
                &[(p vee q) wedge top ] wedge[(p vee neg r) wedge (neg q vee neg r)]\
                &(p vee q) wedge [(p vee neg r) wedge (neg q vee neg r)].
                end{align*}

                Now overall we have ${(p vee q) wedge [(p vee neg r) wedge (neg q vee neg r)]}vee (neg p vee r)$. If we distribute the $vee$ over the $wedge$, we get $[(p vee q) vee ( neg p vee r)]wedge{[(p vee neg r) wedge (neg q vee neg r)]vee(neg p vee r)}$. Focusing on the first half, you can manipulate $(p vee q) vee ( neg p vee r)$ to get $top$ by shuffling parentheses around to get $(p vee neg p) vee (q vee r)$ (I'll leave that to you).



                So we are left with $[(p vee neg r) wedge (neg q vee neg r)] vee (neg p vee r)$. Again let's distribute the $vee$ over the $wedge$:
                begin{align*}
                &[(p vee neg r) vee (neg p vee r)] wedge [(neg q vee neg r) vee (neg p vee r)].
                end{align*}

                Again both halves of this can be manipulated to get $top$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Expand the expression $(p wedge neg q) vee (q wedge neg r)$ by distributing the $vee$ over the $wedge$:
                  begin{align*}
                  &(p wedge neg q) vee (q wedge neg r)\
                  &[(p wedge neg q) vee q] wedge [(p wedge neg q) vee neg r]\
                  &[(p vee q ) wedge (neg q vee q)] wedge [(p vee neg r ) wedge (neg q vee neg r)]\
                  &[(p vee q) wedge top ] wedge[(p vee neg r) wedge (neg q vee neg r)]\
                  &(p vee q) wedge [(p vee neg r) wedge (neg q vee neg r)].
                  end{align*}

                  Now overall we have ${(p vee q) wedge [(p vee neg r) wedge (neg q vee neg r)]}vee (neg p vee r)$. If we distribute the $vee$ over the $wedge$, we get $[(p vee q) vee ( neg p vee r)]wedge{[(p vee neg r) wedge (neg q vee neg r)]vee(neg p vee r)}$. Focusing on the first half, you can manipulate $(p vee q) vee ( neg p vee r)$ to get $top$ by shuffling parentheses around to get $(p vee neg p) vee (q vee r)$ (I'll leave that to you).



                  So we are left with $[(p vee neg r) wedge (neg q vee neg r)] vee (neg p vee r)$. Again let's distribute the $vee$ over the $wedge$:
                  begin{align*}
                  &[(p vee neg r) vee (neg p vee r)] wedge [(neg q vee neg r) vee (neg p vee r)].
                  end{align*}

                  Again both halves of this can be manipulated to get $top$.






                  share|cite|improve this answer









                  $endgroup$



                  Expand the expression $(p wedge neg q) vee (q wedge neg r)$ by distributing the $vee$ over the $wedge$:
                  begin{align*}
                  &(p wedge neg q) vee (q wedge neg r)\
                  &[(p wedge neg q) vee q] wedge [(p wedge neg q) vee neg r]\
                  &[(p vee q ) wedge (neg q vee q)] wedge [(p vee neg r ) wedge (neg q vee neg r)]\
                  &[(p vee q) wedge top ] wedge[(p vee neg r) wedge (neg q vee neg r)]\
                  &(p vee q) wedge [(p vee neg r) wedge (neg q vee neg r)].
                  end{align*}

                  Now overall we have ${(p vee q) wedge [(p vee neg r) wedge (neg q vee neg r)]}vee (neg p vee r)$. If we distribute the $vee$ over the $wedge$, we get $[(p vee q) vee ( neg p vee r)]wedge{[(p vee neg r) wedge (neg q vee neg r)]vee(neg p vee r)}$. Focusing on the first half, you can manipulate $(p vee q) vee ( neg p vee r)$ to get $top$ by shuffling parentheses around to get $(p vee neg p) vee (q vee r)$ (I'll leave that to you).



                  So we are left with $[(p vee neg r) wedge (neg q vee neg r)] vee (neg p vee r)$. Again let's distribute the $vee$ over the $wedge$:
                  begin{align*}
                  &[(p vee neg r) vee (neg p vee r)] wedge [(neg q vee neg r) vee (neg p vee r)].
                  end{align*}

                  Again both halves of this can be manipulated to get $top$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 4 hours ago









                  kccukccu

                  9,80811128




                  9,80811128























                      1












                      $begingroup$


                      I can't seem to figure out what comes after this step. Can someone help me?




                      Yes.



                      $$begin{align}vdotsquad\iff&~big((p land lnot q) lor (qlandlnot r)big) lor (lnot plor r)
                      \[1ex]iff &~big(lnot plor (pland lnot q)big)lor big(rlor (lnot rland q)big)&quadtextsf{(Commutation and Association)}\vdotsquadend{align}$$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$


                        I can't seem to figure out what comes after this step. Can someone help me?




                        Yes.



                        $$begin{align}vdotsquad\iff&~big((p land lnot q) lor (qlandlnot r)big) lor (lnot plor r)
                        \[1ex]iff &~big(lnot plor (pland lnot q)big)lor big(rlor (lnot rland q)big)&quadtextsf{(Commutation and Association)}\vdotsquadend{align}$$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$


                          I can't seem to figure out what comes after this step. Can someone help me?




                          Yes.



                          $$begin{align}vdotsquad\iff&~big((p land lnot q) lor (qlandlnot r)big) lor (lnot plor r)
                          \[1ex]iff &~big(lnot plor (pland lnot q)big)lor big(rlor (lnot rland q)big)&quadtextsf{(Commutation and Association)}\vdotsquadend{align}$$






                          share|cite|improve this answer









                          $endgroup$




                          I can't seem to figure out what comes after this step. Can someone help me?




                          Yes.



                          $$begin{align}vdotsquad\iff&~big((p land lnot q) lor (qlandlnot r)big) lor (lnot plor r)
                          \[1ex]iff &~big(lnot plor (pland lnot q)big)lor big(rlor (lnot rland q)big)&quadtextsf{(Commutation and Association)}\vdotsquadend{align}$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 3 hours ago









                          Graham KempGraham Kemp

                          85.1k43378




                          85.1k43378























                              1












                              $begingroup$

                              You can use Double Distribution to get
                              $$[(p lor q)land(q lor lnot q)land(lnot q lor lnot r)land(p lor lnot r)]lor (lnot p lor r)$$
                              $q lor lnot q$ is a Tautology so this becomes
                              $$[(p lor q)land(lnot q lor lnot r)land(p lor lnot r)]lor (lnot p lor r)$$
                              which by distribution is
                              $$[(p lor q)land[lnot rland(p lor lnot q)]]lor (lnot p lor r)$$
                              Association gives
                              $$[lnot r land [(p lor q)land(p lor lnot q)]]lor (lnot p lor r)$$
                              Distribution again gives
                              $$[lnot r land [p lor(q land lnot q)]]lor (lnot p lor r)$$
                              $q land lnot q$ is a contradiction so this becomes
                              $$(lnot r land p)lor (lnot p lor r)$$
                              Which by DeMorgan's Law is
                              $$(lnot r land p)lor lnot(lnot r land p)$$
                              Which is a tautology.






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                Thanks for the help!!
                                $endgroup$
                                – Nev
                                2 hours ago
















                              1












                              $begingroup$

                              You can use Double Distribution to get
                              $$[(p lor q)land(q lor lnot q)land(lnot q lor lnot r)land(p lor lnot r)]lor (lnot p lor r)$$
                              $q lor lnot q$ is a Tautology so this becomes
                              $$[(p lor q)land(lnot q lor lnot r)land(p lor lnot r)]lor (lnot p lor r)$$
                              which by distribution is
                              $$[(p lor q)land[lnot rland(p lor lnot q)]]lor (lnot p lor r)$$
                              Association gives
                              $$[lnot r land [(p lor q)land(p lor lnot q)]]lor (lnot p lor r)$$
                              Distribution again gives
                              $$[lnot r land [p lor(q land lnot q)]]lor (lnot p lor r)$$
                              $q land lnot q$ is a contradiction so this becomes
                              $$(lnot r land p)lor (lnot p lor r)$$
                              Which by DeMorgan's Law is
                              $$(lnot r land p)lor lnot(lnot r land p)$$
                              Which is a tautology.






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                Thanks for the help!!
                                $endgroup$
                                – Nev
                                2 hours ago














                              1












                              1








                              1





                              $begingroup$

                              You can use Double Distribution to get
                              $$[(p lor q)land(q lor lnot q)land(lnot q lor lnot r)land(p lor lnot r)]lor (lnot p lor r)$$
                              $q lor lnot q$ is a Tautology so this becomes
                              $$[(p lor q)land(lnot q lor lnot r)land(p lor lnot r)]lor (lnot p lor r)$$
                              which by distribution is
                              $$[(p lor q)land[lnot rland(p lor lnot q)]]lor (lnot p lor r)$$
                              Association gives
                              $$[lnot r land [(p lor q)land(p lor lnot q)]]lor (lnot p lor r)$$
                              Distribution again gives
                              $$[lnot r land [p lor(q land lnot q)]]lor (lnot p lor r)$$
                              $q land lnot q$ is a contradiction so this becomes
                              $$(lnot r land p)lor (lnot p lor r)$$
                              Which by DeMorgan's Law is
                              $$(lnot r land p)lor lnot(lnot r land p)$$
                              Which is a tautology.






                              share|cite|improve this answer









                              $endgroup$



                              You can use Double Distribution to get
                              $$[(p lor q)land(q lor lnot q)land(lnot q lor lnot r)land(p lor lnot r)]lor (lnot p lor r)$$
                              $q lor lnot q$ is a Tautology so this becomes
                              $$[(p lor q)land(lnot q lor lnot r)land(p lor lnot r)]lor (lnot p lor r)$$
                              which by distribution is
                              $$[(p lor q)land[lnot rland(p lor lnot q)]]lor (lnot p lor r)$$
                              Association gives
                              $$[lnot r land [(p lor q)land(p lor lnot q)]]lor (lnot p lor r)$$
                              Distribution again gives
                              $$[lnot r land [p lor(q land lnot q)]]lor (lnot p lor r)$$
                              $q land lnot q$ is a contradiction so this becomes
                              $$(lnot r land p)lor (lnot p lor r)$$
                              Which by DeMorgan's Law is
                              $$(lnot r land p)lor lnot(lnot r land p)$$
                              Which is a tautology.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 3 hours ago









                              Erik ParkinsonErik Parkinson

                              8899




                              8899












                              • $begingroup$
                                Thanks for the help!!
                                $endgroup$
                                – Nev
                                2 hours ago


















                              • $begingroup$
                                Thanks for the help!!
                                $endgroup$
                                – Nev
                                2 hours ago
















                              $begingroup$
                              Thanks for the help!!
                              $endgroup$
                              – Nev
                              2 hours ago




                              $begingroup$
                              Thanks for the help!!
                              $endgroup$
                              – Nev
                              2 hours ago










                              Nev is a new contributor. Be nice, and check out our Code of Conduct.










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                              Nev is a new contributor. Be nice, and check out our Code of Conduct.












                              Nev is a new contributor. Be nice, and check out our Code of Conduct.
















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