Mixing
$(P land neg Q) lor P equiv P$ How is this proved using theorems?
I have tried my best by using all possible ways but failed!
Commutative laws: $p land q equiv q land p$ and $p lor q equiv q lor p$
Associative laws: $(p land q) land r equiv p land (q land r)$ and $(p lor q) lor r equiv p lor (q lor r)$
Distributive laws: $p land (q lor r) equiv (p land q) lor (p land r)$ and $p lor (q land r) equiv (p lor q) land (p lor r)$
Identity laws: $p land top equiv p$ and $p lor bot equiv p$
Negation laws: $p lor neg p equiv top$ and $p land neg p equiv bot$
Double negative law: $neg (neg p) equiv p$
Idempotent laws: $p land p equiv p$ and $p lor p equiv p$
Universal bound laws: $p lor top equiv top$ and $p land bot equiv bot$
De Morgan’s laws: $neg (p land q) equiv neg p lor neg q$ and $neg (p lor q) equiv neg p land neg q$
Absorption laws: $p lor (p land q) equiv p$ and $p land (p lor q) equiv p$
Negations of $top$ and $bot$: $neg top equiv bot$ and $neg bot equiv top$
discrete-mathematics
edited Dec 31 '18 at 20:00
amWhy
192k28225439
asked Dec 31 '18 at 18:28
Waqad ArshadWaqad Arshad
31
add a comment |
I have tried my best by using all possible ways but failed!
Commutative laws: $p land q equiv q land p$ and $p lor q equiv q lor p$
Associative laws: $(p land q) land r equiv p land (q land r)$ and $(p lor q) lor r equiv p lor (q lor r)$
Distributive laws: $p land (q lor r) equiv (p land q) lor (p land r)$ and $p lor (q land r) equiv (p lor q) land (p lor r)$
Identity laws: $p land top equiv p$ and $p lor bot equiv p$
Negation laws: $p lor neg p equiv top$ and $p land neg p equiv bot$
Double negative law: $neg (neg p) equiv p$
Idempotent laws: $p land p equiv p$ and $p lor p equiv p$
Universal bound laws: $p lor top equiv top$ and $p land bot equiv bot$
De Morgan’s laws: $neg (p land q) equiv neg p lor neg q$ and $neg (p lor q) equiv neg p land neg q$
Absorption laws: $p lor (p land q) equiv p$ and $p land (p lor q) equiv p$
Negations of $top$ and $bot$: $neg top equiv bot$ and $neg bot equiv top$
discrete-mathematics
edited Dec 31 '18 at 20:00
amWhy
192k28225439
asked Dec 31 '18 at 18:28
Waqad ArshadWaqad Arshad
31
Whoops, misread it. Thanks @MauroALLEGRANZA.
– mweiss
Dec 31 '18 at 18:59
but I asked how can we prove it using theorems?@MauroALLEGRANZA
– Waqad Arshad
Dec 31 '18 at 19:19
add a comment |
I have tried my best by using all possible ways but failed!
Commutative laws: $p land q equiv q land p$ and $p lor q equiv q lor p$
Associative laws: $(p land q) land r equiv p land (q land r)$ and $(p lor q) lor r equiv p lor (q lor r)$
Distributive laws: $p land (q lor r) equiv (p land q) lor (p land r)$ and $p lor (q land r) equiv (p lor q) land (p lor r)$
Identity laws: $p land top equiv p$ and $p lor bot equiv p$
Negation laws: $p lor neg p equiv top$ and $p land neg p equiv bot$
Double negative law: $neg (neg p) equiv p$
Idempotent laws: $p land p equiv p$ and $p lor p equiv p$
Universal bound laws: $p lor top equiv top$ and $p land bot equiv bot$
De Morgan’s laws: $neg (p land q) equiv neg p lor neg q$ and $neg (p lor q) equiv neg p land neg q$
Absorption laws: $p lor (p land q) equiv p$ and $p land (p lor q) equiv p$
Negations of $top$ and $bot$: $neg top equiv bot$ and $neg bot equiv top$
discrete-mathematics
edited Dec 31 '18 at 20:00
amWhy
192k28225439
asked Dec 31 '18 at 18:28
Waqad ArshadWaqad Arshad
31
I have tried my best by using all possible ways but failed!
Commutative laws: $p land q equiv q land p$ and $p lor q equiv q lor p$
Associative laws: $(p land q) land r equiv p land (q land r)$ and $(p lor q) lor r equiv p lor (q lor r)$
Distributive laws: $p land (q lor r) equiv (p land q) lor (p land r)$ and $p lor (q land r) equiv (p lor q) land (p lor r)$
Identity laws: $p land top equiv p$ and $p lor bot equiv p$
Negation laws: $p lor neg p equiv top$ and $p land neg p equiv bot$
Double negative law: $neg (neg p) equiv p$
Idempotent laws: $p land p equiv p$ and $p lor p equiv p$
Universal bound laws: $p lor top equiv top$ and $p land bot equiv bot$
De Morgan’s laws: $neg (p land q) equiv neg p lor neg q$ and $neg (p lor q) equiv neg p land neg q$
Absorption laws: $p lor (p land q) equiv p$ and $p land (p lor q) equiv p$
Negations of $top$ and $bot$: $neg top equiv bot$ and $neg bot equiv top$
discrete-mathematics
discrete-mathematics
edited Dec 31 '18 at 20:00
amWhy
192k28225439
asked Dec 31 '18 at 18:28
Waqad ArshadWaqad Arshad
31
edited Dec 31 '18 at 20:00
amWhy
192k28225439
asked Dec 31 '18 at 18:28
Waqad ArshadWaqad Arshad
31
edited Dec 31 '18 at 20:00
amWhy
192k28225439
edited Dec 31 '18 at 20:00
amWhy
192k28225439
edited Dec 31 '18 at 20:00
amWhy
192k28225439
192k28225439
asked Dec 31 '18 at 18:28
Waqad ArshadWaqad Arshad
31
asked Dec 31 '18 at 18:28
Waqad ArshadWaqad Arshad
31
asked Dec 31 '18 at 18:28
Waqad ArshadWaqad Arshad
31
31
Whoops, misread it. Thanks @MauroALLEGRANZA.
– mweiss
Dec 31 '18 at 18:59
but I asked how can we prove it using theorems?@MauroALLEGRANZA
– Waqad Arshad
Dec 31 '18 at 19:19
add a comment |
Whoops, misread it. Thanks @MauroALLEGRANZA.
– mweiss
Dec 31 '18 at 18:59
but I asked how can we prove it using theorems?@MauroALLEGRANZA
– Waqad Arshad
Dec 31 '18 at 19:19
Whoops, misread it. Thanks @MauroALLEGRANZA.
– mweiss
Dec 31 '18 at 18:59
Whoops, misread it. Thanks @MauroALLEGRANZA.
– mweiss
Dec 31 '18 at 18:59
but I asked how can we prove it using theorems?@MauroALLEGRANZA
– Waqad Arshad
Dec 31 '18 at 19:19
but I asked how can we prove it using theorems?@MauroALLEGRANZA
– Waqad Arshad
Dec 31 '18 at 19:19
add a comment |
1 Answer
1
active
oldest
votes
It can be proved from the absorption law that you list with a variable substitution. Let $Q'=neg Q$. Then by the absorption law, $(P wedge Q') vee P equiv P$.
answered Dec 31 '18 at 18:41
CyborgOctopusCyborgOctopus
685
would it be correct?
– Waqad Arshad
Dec 31 '18 at 19:15
@WaqadArshad yes. It follows pretty directly from the law that you listed and like someone else mentioned, you can check it with a truth table.
– CyborgOctopus
Dec 31 '18 at 19:18
I know that this statement is true. all I am asking is that is it alright to write it this way?
– Waqad Arshad
Dec 31 '18 at 19:41
You might want to substitute ~Q back in to rewrite it in its original form.
– CyborgOctopus
Dec 31 '18 at 19:45
Thanks a lot @CyborgOctopus
– Waqad Arshad
Dec 31 '18 at 19:47
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
It can be proved from the absorption law that you list with a variable substitution. Let $Q'=neg Q$. Then by the absorption law, $(P wedge Q') vee P equiv P$.
answered Dec 31 '18 at 18:41
CyborgOctopusCyborgOctopus
685
would it be correct?
– Waqad Arshad
Dec 31 '18 at 19:15
@WaqadArshad yes. It follows pretty directly from the law that you listed and like someone else mentioned, you can check it with a truth table.
– CyborgOctopus
Dec 31 '18 at 19:18
I know that this statement is true. all I am asking is that is it alright to write it this way?
– Waqad Arshad
Dec 31 '18 at 19:41
You might want to substitute ~Q back in to rewrite it in its original form.
– CyborgOctopus
Dec 31 '18 at 19:45
Thanks a lot @CyborgOctopus
– Waqad Arshad
Dec 31 '18 at 19:47
add a comment |
It can be proved from the absorption law that you list with a variable substitution. Let $Q'=neg Q$. Then by the absorption law, $(P wedge Q') vee P equiv P$.
answered Dec 31 '18 at 18:41
CyborgOctopusCyborgOctopus
685
would it be correct?
– Waqad Arshad
Dec 31 '18 at 19:15
@WaqadArshad yes. It follows pretty directly from the law that you listed and like someone else mentioned, you can check it with a truth table.
– CyborgOctopus
Dec 31 '18 at 19:18
I know that this statement is true. all I am asking is that is it alright to write it this way?
– Waqad Arshad
Dec 31 '18 at 19:41
You might want to substitute ~Q back in to rewrite it in its original form.
– CyborgOctopus
Dec 31 '18 at 19:45
Thanks a lot @CyborgOctopus
– Waqad Arshad
Dec 31 '18 at 19:47
add a comment |
It can be proved from the absorption law that you list with a variable substitution. Let $Q'=neg Q$. Then by the absorption law, $(P wedge Q') vee P equiv P$.
answered Dec 31 '18 at 18:41
CyborgOctopusCyborgOctopus
685
It can be proved from the absorption law that you list with a variable substitution. Let $Q'=neg Q$. Then by the absorption law, $(P wedge Q') vee P equiv P$.
answered Dec 31 '18 at 18:41
CyborgOctopusCyborgOctopus
685
answered Dec 31 '18 at 18:41
CyborgOctopusCyborgOctopus
685
answered Dec 31 '18 at 18:41
CyborgOctopusCyborgOctopus
685
answered Dec 31 '18 at 18:41
CyborgOctopusCyborgOctopus
685
685
would it be correct?
– Waqad Arshad
Dec 31 '18 at 19:15
@WaqadArshad yes. It follows pretty directly from the law that you listed and like someone else mentioned, you can check it with a truth table.
– CyborgOctopus
Dec 31 '18 at 19:18
I know that this statement is true. all I am asking is that is it alright to write it this way?
– Waqad Arshad
Dec 31 '18 at 19:41
You might want to substitute ~Q back in to rewrite it in its original form.
– CyborgOctopus
Dec 31 '18 at 19:45
Thanks a lot @CyborgOctopus
– Waqad Arshad
Dec 31 '18 at 19:47
add a comment |
would it be correct?
– Waqad Arshad
Dec 31 '18 at 19:15
@WaqadArshad yes. It follows pretty directly from the law that you listed and like someone else mentioned, you can check it with a truth table.
– CyborgOctopus
Dec 31 '18 at 19:18
I know that this statement is true. all I am asking is that is it alright to write it this way?
– Waqad Arshad
Dec 31 '18 at 19:41
You might want to substitute ~Q back in to rewrite it in its original form.
– CyborgOctopus
Dec 31 '18 at 19:45
Thanks a lot @CyborgOctopus
– Waqad Arshad
Dec 31 '18 at 19:47
would it be correct?
– Waqad Arshad
Dec 31 '18 at 19:15
would it be correct?
– Waqad Arshad
Dec 31 '18 at 19:15
@WaqadArshad yes. It follows pretty directly from the law that you listed and like someone else mentioned, you can check it with a truth table.
– CyborgOctopus
Dec 31 '18 at 19:18
@WaqadArshad yes. It follows pretty directly from the law that you listed and like someone else mentioned, you can check it with a truth table.
– CyborgOctopus
Dec 31 '18 at 19:18
I know that this statement is true. all I am asking is that is it alright to write it this way?
– Waqad Arshad
Dec 31 '18 at 19:41
I know that this statement is true. all I am asking is that is it alright to write it this way?
– Waqad Arshad
Dec 31 '18 at 19:41
You might want to substitute ~Q back in to rewrite it in its original form.
– CyborgOctopus
Dec 31 '18 at 19:45
You might want to substitute ~Q back in to rewrite it in its original form.
– CyborgOctopus
Dec 31 '18 at 19:45
Thanks a lot @CyborgOctopus
– Waqad Arshad
Dec 31 '18 at 19:47
Thanks a lot @CyborgOctopus
– Waqad Arshad
Dec 31 '18 at 19:47
add a comment |
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Whoops, misread it. Thanks @MauroALLEGRANZA.
– mweiss
Dec 31 '18 at 18:59
but I asked how can we prove it using theorems?@MauroALLEGRANZA
– Waqad Arshad
Dec 31 '18 at 19:19