I'm trying to prove P→(Q∧R) from the following premises: (P→Q)∧R, (P∧R)→S, ¬S [closed]
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I made a truth table and found that the conclusion is valid. But, I need help constructing a formal proof.
discrete-mathematics natural-deduction
asked Nov 16 at 5:24
Orlando Piedrahita
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closed as off-topic by Leucippus, Key Flex, José Carlos Santos, Shailesh, Tom-Tom Nov 16 at 8:41
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I made a truth table and found that the conclusion is valid. But, I need help constructing a formal proof.
discrete-mathematics natural-deduction
asked Nov 16 at 5:24
Orlando Piedrahita
11
closed as off-topic by Leucippus, Key Flex, José Carlos Santos, Shailesh, Tom-Tom Nov 16 at 8:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Key Flex, José Carlos Santos, Shailesh, Tom-Tom
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I made a truth table and found that the conclusion is valid. But, I need help constructing a formal proof.
discrete-mathematics natural-deduction
asked Nov 16 at 5:24
Orlando Piedrahita
11
I made a truth table and found that the conclusion is valid. But, I need help constructing a formal proof.
discrete-mathematics natural-deduction
discrete-mathematics natural-deduction
asked Nov 16 at 5:24
Orlando Piedrahita
11
asked Nov 16 at 5:24
Orlando Piedrahita
11
asked Nov 16 at 5:24
Orlando Piedrahita
11
asked Nov 16 at 5:24
Orlando Piedrahita
11
asked Nov 16 at 5:24
Orlando Piedrahita
11
11
closed as off-topic by Leucippus, Key Flex, José Carlos Santos, Shailesh, Tom-Tom Nov 16 at 8:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Key Flex, José Carlos Santos, Shailesh, Tom-Tom
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Leucippus, Key Flex, José Carlos Santos, Shailesh, Tom-Tom Nov 16 at 8:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Key Flex, José Carlos Santos, Shailesh, Tom-Tom
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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From your question, I assume you are trying to prove:
$$((P implies Q) land R) land (P land R implies S) land lnot S) implies (P implies (Q land R))$$
Whenever you want to prove an implication you assume the first part.
Now assume P.
Now you are given $$(P land R) implies S$$
You have P because you assumed and you have R because of your givens. Now you S. But you also have lnot S. This gives you a $bot$. From a contraction you can assume anything. The rest is minor details.
answered Nov 16 at 6:07
Fefnir Wilhelm
62
1
Doesn't $((Pimplies Q)land R)$ by itself imply $(Pimplies(Qland R))$?
– bof
Nov 16 at 6:13
Ah yes, I do believe that is better way of doing the problem! because if you know $P$ conclude $$Q$$ and it must follow $$Q land R$$
– Fefnir Wilhelm
Nov 16 at 16:42
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
From your question, I assume you are trying to prove:
$$((P implies Q) land R) land (P land R implies S) land lnot S) implies (P implies (Q land R))$$
Whenever you want to prove an implication you assume the first part.
Now assume P.
Now you are given $$(P land R) implies S$$
You have P because you assumed and you have R because of your givens. Now you S. But you also have lnot S. This gives you a $bot$. From a contraction you can assume anything. The rest is minor details.
answered Nov 16 at 6:07
Fefnir Wilhelm
62
1
Doesn't $((Pimplies Q)land R)$ by itself imply $(Pimplies(Qland R))$?
– bof
Nov 16 at 6:13
Ah yes, I do believe that is better way of doing the problem! because if you know $P$ conclude $$Q$$ and it must follow $$Q land R$$
– Fefnir Wilhelm
Nov 16 at 16:42
add a comment |
up vote
0
down vote
From your question, I assume you are trying to prove:
$$((P implies Q) land R) land (P land R implies S) land lnot S) implies (P implies (Q land R))$$
Whenever you want to prove an implication you assume the first part.
Now assume P.
Now you are given $$(P land R) implies S$$
You have P because you assumed and you have R because of your givens. Now you S. But you also have lnot S. This gives you a $bot$. From a contraction you can assume anything. The rest is minor details.
answered Nov 16 at 6:07
Fefnir Wilhelm
62
1
Doesn't $((Pimplies Q)land R)$ by itself imply $(Pimplies(Qland R))$?
– bof
Nov 16 at 6:13
Ah yes, I do believe that is better way of doing the problem! because if you know $P$ conclude $$Q$$ and it must follow $$Q land R$$
– Fefnir Wilhelm
Nov 16 at 16:42
add a comment |
up vote
0
down vote
up vote
0
down vote
From your question, I assume you are trying to prove:
$$((P implies Q) land R) land (P land R implies S) land lnot S) implies (P implies (Q land R))$$
Whenever you want to prove an implication you assume the first part.
Now assume P.
Now you are given $$(P land R) implies S$$
You have P because you assumed and you have R because of your givens. Now you S. But you also have lnot S. This gives you a $bot$. From a contraction you can assume anything. The rest is minor details.
answered Nov 16 at 6:07
Fefnir Wilhelm
62
From your question, I assume you are trying to prove:
$$((P implies Q) land R) land (P land R implies S) land lnot S) implies (P implies (Q land R))$$
Whenever you want to prove an implication you assume the first part.
Now assume P.
Now you are given $$(P land R) implies S$$
You have P because you assumed and you have R because of your givens. Now you S. But you also have lnot S. This gives you a $bot$. From a contraction you can assume anything. The rest is minor details.
answered Nov 16 at 6:07
Fefnir Wilhelm
62
answered Nov 16 at 6:07
Fefnir Wilhelm
62
answered Nov 16 at 6:07
Fefnir Wilhelm
62
answered Nov 16 at 6:07
Fefnir Wilhelm
62
62
1
Doesn't $((Pimplies Q)land R)$ by itself imply $(Pimplies(Qland R))$?
– bof
Nov 16 at 6:13
Ah yes, I do believe that is better way of doing the problem! because if you know $P$ conclude $$Q$$ and it must follow $$Q land R$$
– Fefnir Wilhelm
Nov 16 at 16:42
add a comment |
1
Doesn't $((Pimplies Q)land R)$ by itself imply $(Pimplies(Qland R))$?
– bof
Nov 16 at 6:13
Ah yes, I do believe that is better way of doing the problem! because if you know $P$ conclude $$Q$$ and it must follow $$Q land R$$
– Fefnir Wilhelm
Nov 16 at 16:42
1
1
Doesn't $((Pimplies Q)land R)$ by itself imply $(Pimplies(Qland R))$?
– bof
Nov 16 at 6:13
Doesn't $((Pimplies Q)land R)$ by itself imply $(Pimplies(Qland R))$?
– bof
Nov 16 at 6:13
Ah yes, I do believe that is better way of doing the problem! because if you know $P$ conclude $$Q$$ and it must follow $$Q land R$$
– Fefnir Wilhelm
Nov 16 at 16:42
Ah yes, I do believe that is better way of doing the problem! because if you know $P$ conclude $$Q$$ and it must follow $$Q land R$$
– Fefnir Wilhelm
Nov 16 at 16:42
add a comment |
