Proving Logical Equivalence With Basic Laws of Logic
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I have to prove the following equivalence:
$(X rightarrow (Y lor Z)) lor (X leftrightarrow (Y land Z)) equiv neg(X land neg Y land neg Z)$
This is how far I've come:
$equiv (neg X lor (Y lor Z)) lor (X leftrightarrow (Y land Z))$
$equiv (neg X lor (Y lor Z)) lor ((X rightarrow (Y land Z)) land ((Y land Z) rightarrow X) )$
$equiv (neg X lor (Y lor Z)) lor ((neg X lor (Y land Z)) land (neg (Y land Z) lor X) )$
$equiv (neg X lor (Y lor Z)) lor ((neg X lor (Y land Z)) land (X lor neg (Y land Z)))$
$equiv neg [neg(neg X lor (Y lor Z)) land neg((neg X lor (Y land Z)) land (X lor neg (Y land Z)))]$
$equiv neg[( X land (neg Y land neg Z)) land neg((neg X lor (Y land Z)) land (X lor neg (Y land Z)))]$
$equiv neg[(X land (neg Y land neg Z)) land (neg(neg X lor (Y land Z)) lor neg(X lor neg (Y land Z)))]$
$equiv neg[(X land ( neg Y land neg Z)) land ((X land neg(Y land Z)) lor neg(X land neg (Y land Z))]$
$equiv neg[(X land (neg Y land neg Z)) land ((X land neg(Y land Z)) lor (neg X land (Y land Z)))]$
logic
asked Nov 17 at 16:51
Marina A.
1
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up vote
0
down vote
favorite
I have to prove the following equivalence:
$(X rightarrow (Y lor Z)) lor (X leftrightarrow (Y land Z)) equiv neg(X land neg Y land neg Z)$
This is how far I've come:
$equiv (neg X lor (Y lor Z)) lor (X leftrightarrow (Y land Z))$
$equiv (neg X lor (Y lor Z)) lor ((X rightarrow (Y land Z)) land ((Y land Z) rightarrow X) )$
$equiv (neg X lor (Y lor Z)) lor ((neg X lor (Y land Z)) land (neg (Y land Z) lor X) )$
$equiv (neg X lor (Y lor Z)) lor ((neg X lor (Y land Z)) land (X lor neg (Y land Z)))$
$equiv neg [neg(neg X lor (Y lor Z)) land neg((neg X lor (Y land Z)) land (X lor neg (Y land Z)))]$
$equiv neg[( X land (neg Y land neg Z)) land neg((neg X lor (Y land Z)) land (X lor neg (Y land Z)))]$
$equiv neg[(X land (neg Y land neg Z)) land (neg(neg X lor (Y land Z)) lor neg(X lor neg (Y land Z)))]$
$equiv neg[(X land ( neg Y land neg Z)) land ((X land neg(Y land Z)) lor neg(X land neg (Y land Z))]$
$equiv neg[(X land (neg Y land neg Z)) land ((X land neg(Y land Z)) lor (neg X land (Y land Z)))]$
logic
asked Nov 17 at 16:51
Marina A.
1
Do you require a solution using laaws of logic? Because there are only 3 variables here, so you could just draw a truth table.
– Josh B.
Nov 17 at 17:33
I don't know for sure, because we have always solved it this way. But you are right, a truth table would be way easier.
– Marina A.
Nov 18 at 7:06
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have to prove the following equivalence:
$(X rightarrow (Y lor Z)) lor (X leftrightarrow (Y land Z)) equiv neg(X land neg Y land neg Z)$
This is how far I've come:
$equiv (neg X lor (Y lor Z)) lor (X leftrightarrow (Y land Z))$
$equiv (neg X lor (Y lor Z)) lor ((X rightarrow (Y land Z)) land ((Y land Z) rightarrow X) )$
$equiv (neg X lor (Y lor Z)) lor ((neg X lor (Y land Z)) land (neg (Y land Z) lor X) )$
$equiv (neg X lor (Y lor Z)) lor ((neg X lor (Y land Z)) land (X lor neg (Y land Z)))$
$equiv neg [neg(neg X lor (Y lor Z)) land neg((neg X lor (Y land Z)) land (X lor neg (Y land Z)))]$
$equiv neg[( X land (neg Y land neg Z)) land neg((neg X lor (Y land Z)) land (X lor neg (Y land Z)))]$
$equiv neg[(X land (neg Y land neg Z)) land (neg(neg X lor (Y land Z)) lor neg(X lor neg (Y land Z)))]$
$equiv neg[(X land ( neg Y land neg Z)) land ((X land neg(Y land Z)) lor neg(X land neg (Y land Z))]$
$equiv neg[(X land (neg Y land neg Z)) land ((X land neg(Y land Z)) lor (neg X land (Y land Z)))]$
logic
asked Nov 17 at 16:51
Marina A.
1
I have to prove the following equivalence:
$(X rightarrow (Y lor Z)) lor (X leftrightarrow (Y land Z)) equiv neg(X land neg Y land neg Z)$
This is how far I've come:
$equiv (neg X lor (Y lor Z)) lor (X leftrightarrow (Y land Z))$
$equiv (neg X lor (Y lor Z)) lor ((X rightarrow (Y land Z)) land ((Y land Z) rightarrow X) )$
$equiv (neg X lor (Y lor Z)) lor ((neg X lor (Y land Z)) land (neg (Y land Z) lor X) )$
$equiv (neg X lor (Y lor Z)) lor ((neg X lor (Y land Z)) land (X lor neg (Y land Z)))$
$equiv neg [neg(neg X lor (Y lor Z)) land neg((neg X lor (Y land Z)) land (X lor neg (Y land Z)))]$
$equiv neg[( X land (neg Y land neg Z)) land neg((neg X lor (Y land Z)) land (X lor neg (Y land Z)))]$
$equiv neg[(X land (neg Y land neg Z)) land (neg(neg X lor (Y land Z)) lor neg(X lor neg (Y land Z)))]$
$equiv neg[(X land ( neg Y land neg Z)) land ((X land neg(Y land Z)) lor neg(X land neg (Y land Z))]$
$equiv neg[(X land (neg Y land neg Z)) land ((X land neg(Y land Z)) lor (neg X land (Y land Z)))]$
logic
logic
asked Nov 17 at 16:51
Marina A.
1
asked Nov 17 at 16:51
Marina A.
1
asked Nov 17 at 16:51
Marina A.
1
asked Nov 17 at 16:51
Marina A.
1
asked Nov 17 at 16:51
Marina A.
1
1
Do you require a solution using laaws of logic? Because there are only 3 variables here, so you could just draw a truth table.
– Josh B.
Nov 17 at 17:33
I don't know for sure, because we have always solved it this way. But you are right, a truth table would be way easier.
– Marina A.
Nov 18 at 7:06
add a comment |
Do you require a solution using laaws of logic? Because there are only 3 variables here, so you could just draw a truth table.
– Josh B.
Nov 17 at 17:33
I don't know for sure, because we have always solved it this way. But you are right, a truth table would be way easier.
– Marina A.
Nov 18 at 7:06
Do you require a solution using laaws of logic? Because there are only 3 variables here, so you could just draw a truth table.
– Josh B.
Nov 17 at 17:33
Do you require a solution using laaws of logic? Because there are only 3 variables here, so you could just draw a truth table.
– Josh B.
Nov 17 at 17:33
I don't know for sure, because we have always solved it this way. But you are right, a truth table would be way easier.
– Marina A.
Nov 18 at 7:06
I don't know for sure, because we have always solved it this way. But you are right, a truth table would be way easier.
– Marina A.
Nov 18 at 7:06
add a comment |
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Do you require a solution using laaws of logic? Because there are only 3 variables here, so you could just draw a truth table.
– Josh B.
Nov 17 at 17:33
I don't know for sure, because we have always solved it this way. But you are right, a truth table would be way easier.
– Marina A.
Nov 18 at 7:06