Mixing
∼p∨(∼p∧q)≡∼p∧∼q , prove logical equivalence
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Our teacher gave us this equivalence to prove in our final exam but almost all of us said that it was not provable and we did not solve it and now we asked him (The Teacher) and he says that it is provable. I was unable to prove it even by using truth table and he asked us to prove it by using the theorems for logical equivalence.
So now we have our grades at stake so please solve it so that I may know that either I'm wrong or the Teacher is!
The theorems are:
Commutative laws: p∧q ≡ q∧p , p∨q ≡ q∨p
Associative laws: (p∧q)∧r ≡ p∧(q∧r) , (p∨q)∨r ≡ p∨(q∨r)
Distributive laws: p∧(q∨r) ≡ (p∧q)∨(p∧r), p∨(q∧r) ≡ (p∨q)∧(p∨r)
Identity laws: p∧t ≡ p , p∨c ≡ p
Negation laws: p∨∼p ≡ t , p∧∼p ≡ c
Double negative law: ∼(∼p) ≡ p
Idempotent laws: p∧p ≡ p , p∨p ≡ p
Universal bound laws: p∨t≡t ,p∧c≡c
De Morgan’s laws: ∼(p∧q) ≡ ∼p∨∼q ,∼(p∨q) ≡ ∼p∧∼q
Absorption laws: p∨(p∧q) ≡ p ,p∧(p∨q) ≡ p
Negations of t and c: ∼t ≡ c , ∼c ≡ t
discrete-mathematics logic
asked Jan 12 at 8:19
Waqad ArshadWaqad Arshad
52
$endgroup$
add a comment |
$begingroup$
Our teacher gave us this equivalence to prove in our final exam but almost all of us said that it was not provable and we did not solve it and now we asked him (The Teacher) and he says that it is provable. I was unable to prove it even by using truth table and he asked us to prove it by using the theorems for logical equivalence.
So now we have our grades at stake so please solve it so that I may know that either I'm wrong or the Teacher is!
The theorems are:
Commutative laws: p∧q ≡ q∧p , p∨q ≡ q∨p
Associative laws: (p∧q)∧r ≡ p∧(q∧r) , (p∨q)∨r ≡ p∨(q∨r)
Distributive laws: p∧(q∨r) ≡ (p∧q)∨(p∧r), p∨(q∧r) ≡ (p∨q)∧(p∨r)
Identity laws: p∧t ≡ p , p∨c ≡ p
Negation laws: p∨∼p ≡ t , p∧∼p ≡ c
Double negative law: ∼(∼p) ≡ p
Idempotent laws: p∧p ≡ p , p∨p ≡ p
Universal bound laws: p∨t≡t ,p∧c≡c
De Morgan’s laws: ∼(p∧q) ≡ ∼p∨∼q ,∼(p∨q) ≡ ∼p∧∼q
Absorption laws: p∨(p∧q) ≡ p ,p∧(p∨q) ≡ p
Negations of t and c: ∼t ≡ c , ∼c ≡ t
discrete-mathematics logic
asked Jan 12 at 8:19
Waqad ArshadWaqad Arshad
52
$endgroup$
$begingroup$
Well, thanks a lot to everyone for clarifying this point. in fact, I did the same in my paper. I proved that it was not holding logical equivalence and then proved that it was equivalent to (not)p. But the teacher still says that his question is correct and it holds logical equivalence.
$endgroup$
– Waqad Arshad
Jan 12 at 8:55
add a comment |
$begingroup$
Our teacher gave us this equivalence to prove in our final exam but almost all of us said that it was not provable and we did not solve it and now we asked him (The Teacher) and he says that it is provable. I was unable to prove it even by using truth table and he asked us to prove it by using the theorems for logical equivalence.
So now we have our grades at stake so please solve it so that I may know that either I'm wrong or the Teacher is!
The theorems are:
Commutative laws: p∧q ≡ q∧p , p∨q ≡ q∨p
Associative laws: (p∧q)∧r ≡ p∧(q∧r) , (p∨q)∨r ≡ p∨(q∨r)
Distributive laws: p∧(q∨r) ≡ (p∧q)∨(p∧r), p∨(q∧r) ≡ (p∨q)∧(p∨r)
Identity laws: p∧t ≡ p , p∨c ≡ p
Negation laws: p∨∼p ≡ t , p∧∼p ≡ c
Double negative law: ∼(∼p) ≡ p
Idempotent laws: p∧p ≡ p , p∨p ≡ p
Universal bound laws: p∨t≡t ,p∧c≡c
De Morgan’s laws: ∼(p∧q) ≡ ∼p∨∼q ,∼(p∨q) ≡ ∼p∧∼q
Absorption laws: p∨(p∧q) ≡ p ,p∧(p∨q) ≡ p
Negations of t and c: ∼t ≡ c , ∼c ≡ t
discrete-mathematics logic
asked Jan 12 at 8:19
Waqad ArshadWaqad Arshad
52
$endgroup$
Our teacher gave us this equivalence to prove in our final exam but almost all of us said that it was not provable and we did not solve it and now we asked him (The Teacher) and he says that it is provable. I was unable to prove it even by using truth table and he asked us to prove it by using the theorems for logical equivalence.
So now we have our grades at stake so please solve it so that I may know that either I'm wrong or the Teacher is!
The theorems are:
Commutative laws: p∧q ≡ q∧p , p∨q ≡ q∨p
Associative laws: (p∧q)∧r ≡ p∧(q∧r) , (p∨q)∨r ≡ p∨(q∨r)
Distributive laws: p∧(q∨r) ≡ (p∧q)∨(p∧r), p∨(q∧r) ≡ (p∨q)∧(p∨r)
Identity laws: p∧t ≡ p , p∨c ≡ p
Negation laws: p∨∼p ≡ t , p∧∼p ≡ c
Double negative law: ∼(∼p) ≡ p
Idempotent laws: p∧p ≡ p , p∨p ≡ p
Universal bound laws: p∨t≡t ,p∧c≡c
De Morgan’s laws: ∼(p∧q) ≡ ∼p∨∼q ,∼(p∨q) ≡ ∼p∧∼q
Absorption laws: p∨(p∧q) ≡ p ,p∧(p∨q) ≡ p
Negations of t and c: ∼t ≡ c , ∼c ≡ t
discrete-mathematics logic
discrete-mathematics logic
asked Jan 12 at 8:19
Waqad ArshadWaqad Arshad
52
asked Jan 12 at 8:19
Waqad ArshadWaqad Arshad
52
edited Jan 12 at 8:24
Waqad Arshad
asked Jan 12 at 8:19
Waqad ArshadWaqad Arshad
52
asked Jan 12 at 8:19
Waqad ArshadWaqad Arshad
52
asked Jan 12 at 8:19
Waqad ArshadWaqad Arshad
52
52
$begingroup$
Well, thanks a lot to everyone for clarifying this point. in fact, I did the same in my paper. I proved that it was not holding logical equivalence and then proved that it was equivalent to (not)p. But the teacher still says that his question is correct and it holds logical equivalence.
$endgroup$
– Waqad Arshad
Jan 12 at 8:55
add a comment |
$begingroup$
Well, thanks a lot to everyone for clarifying this point. in fact, I did the same in my paper. I proved that it was not holding logical equivalence and then proved that it was equivalent to (not)p. But the teacher still says that his question is correct and it holds logical equivalence.
$endgroup$
– Waqad Arshad
Jan 12 at 8:55
$begingroup$
Well, thanks a lot to everyone for clarifying this point. in fact, I did the same in my paper. I proved that it was not holding logical equivalence and then proved that it was equivalent to (not)p. But the teacher still says that his question is correct and it holds logical equivalence.
$endgroup$
– Waqad Arshad
Jan 12 at 8:55
$begingroup$
Well, thanks a lot to everyone for clarifying this point. in fact, I did the same in my paper. I proved that it was not holding logical equivalence and then proved that it was equivalent to (not)p. But the teacher still says that his question is correct and it holds logical equivalence.
$endgroup$
– Waqad Arshad
Jan 12 at 8:55
add a comment |
4 Answers
4
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I'd say you're right. Using 10, $∼p∨(∼p∧q)≡∼p$, but $∼p not equiv ∼p∧q.$
answered Jan 12 at 8:46
PatricioPatricio
3257
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add a comment |
$begingroup$
If $p$ is false, the left side is true since it's an "or" with one term true (since $p$ false, (not p) is true). But suppose at the same time that $q$ is true. Then the right is false, since it's an "and" with one term false (since $q$ is true, (not q) is false). So it isn't an equivalence.
answered Jan 12 at 8:44
coffeemathcoffeemath
2,8451415
$endgroup$
add a comment |
$begingroup$
If $p$ is false and $q$ is true then $neg pwedgeneg q$ is false and $neg pvee(neg pwedge q)$ is true.
So there no logical equivalence.
answered Jan 12 at 8:46
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
Hint
Yes you are right. This is not an equivalence. In fact $$∼p∨(∼p∧q)≡∼p$$using a truth table.
answered Jan 12 at 8:48
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'd say you're right. Using 10, $∼p∨(∼p∧q)≡∼p$, but $∼p not equiv ∼p∧q.$
answered Jan 12 at 8:46
PatricioPatricio
3257
$endgroup$
add a comment |
$begingroup$
I'd say you're right. Using 10, $∼p∨(∼p∧q)≡∼p$, but $∼p not equiv ∼p∧q.$
answered Jan 12 at 8:46
PatricioPatricio
3257
$endgroup$
add a comment |
$begingroup$
I'd say you're right. Using 10, $∼p∨(∼p∧q)≡∼p$, but $∼p not equiv ∼p∧q.$
answered Jan 12 at 8:46
PatricioPatricio
3257
$endgroup$
I'd say you're right. Using 10, $∼p∨(∼p∧q)≡∼p$, but $∼p not equiv ∼p∧q.$
answered Jan 12 at 8:46
PatricioPatricio
3257
answered Jan 12 at 8:46
PatricioPatricio
3257
answered Jan 12 at 8:46
PatricioPatricio
3257
answered Jan 12 at 8:46
PatricioPatricio
3257
3257
add a comment |
add a comment |
$begingroup$
If $p$ is false, the left side is true since it's an "or" with one term true (since $p$ false, (not p) is true). But suppose at the same time that $q$ is true. Then the right is false, since it's an "and" with one term false (since $q$ is true, (not q) is false). So it isn't an equivalence.
answered Jan 12 at 8:44
coffeemathcoffeemath
2,8451415
$endgroup$
add a comment |
$begingroup$
If $p$ is false, the left side is true since it's an "or" with one term true (since $p$ false, (not p) is true). But suppose at the same time that $q$ is true. Then the right is false, since it's an "and" with one term false (since $q$ is true, (not q) is false). So it isn't an equivalence.
answered Jan 12 at 8:44
coffeemathcoffeemath
2,8451415
$endgroup$
add a comment |
$begingroup$
If $p$ is false, the left side is true since it's an "or" with one term true (since $p$ false, (not p) is true). But suppose at the same time that $q$ is true. Then the right is false, since it's an "and" with one term false (since $q$ is true, (not q) is false). So it isn't an equivalence.
answered Jan 12 at 8:44
coffeemathcoffeemath
2,8451415
$endgroup$
If $p$ is false, the left side is true since it's an "or" with one term true (since $p$ false, (not p) is true). But suppose at the same time that $q$ is true. Then the right is false, since it's an "and" with one term false (since $q$ is true, (not q) is false). So it isn't an equivalence.
answered Jan 12 at 8:44
coffeemathcoffeemath
2,8451415
answered Jan 12 at 8:44
coffeemathcoffeemath
2,8451415
answered Jan 12 at 8:44
coffeemathcoffeemath
2,8451415
answered Jan 12 at 8:44
coffeemathcoffeemath
2,8451415
2,8451415
add a comment |
add a comment |
$begingroup$
If $p$ is false and $q$ is true then $neg pwedgeneg q$ is false and $neg pvee(neg pwedge q)$ is true.
So there no logical equivalence.
answered Jan 12 at 8:46
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
If $p$ is false and $q$ is true then $neg pwedgeneg q$ is false and $neg pvee(neg pwedge q)$ is true.
So there no logical equivalence.
answered Jan 12 at 8:46
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
If $p$ is false and $q$ is true then $neg pwedgeneg q$ is false and $neg pvee(neg pwedge q)$ is true.
So there no logical equivalence.
answered Jan 12 at 8:46
drhabdrhab
101k544130
$endgroup$
If $p$ is false and $q$ is true then $neg pwedgeneg q$ is false and $neg pvee(neg pwedge q)$ is true.
So there no logical equivalence.
answered Jan 12 at 8:46
drhabdrhab
101k544130
answered Jan 12 at 8:46
drhabdrhab
101k544130
answered Jan 12 at 8:46
drhabdrhab
101k544130
answered Jan 12 at 8:46
drhabdrhab
101k544130
101k544130
add a comment |
add a comment |
$begingroup$
Hint
Yes you are right. This is not an equivalence. In fact $$∼p∨(∼p∧q)≡∼p$$using a truth table.
answered Jan 12 at 8:48
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Hint
Yes you are right. This is not an equivalence. In fact $$∼p∨(∼p∧q)≡∼p$$using a truth table.
answered Jan 12 at 8:48
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Hint
Yes you are right. This is not an equivalence. In fact $$∼p∨(∼p∧q)≡∼p$$using a truth table.
answered Jan 12 at 8:48
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
Hint
Yes you are right. This is not an equivalence. In fact $$∼p∨(∼p∧q)≡∼p$$using a truth table.
answered Jan 12 at 8:48
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 8:48
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 8:48
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 8:48
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
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$begingroup$
Well, thanks a lot to everyone for clarifying this point. in fact, I did the same in my paper. I proved that it was not holding logical equivalence and then proved that it was equivalent to (not)p. But the teacher still says that his question is correct and it holds logical equivalence.
$endgroup$
– Waqad Arshad
Jan 12 at 8:55