Mixing
How can I prove the following with natural deduction rules? ¬∀x∃yP(x,y) ⊢ ∃x∀y¬P(x,y)
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I have been stuck with this problem for a long time, I tried reductio ad absurdum and I got the hypothesys [¬∃x∀y¬P(x,y)], then I try to eliminate the negation of the premise, but I have to prove ∀x∃yP(x,y), and after using the introduction of universal quantifier rule, I go again with reductio ad absurdum, gaining a second hypothesis [¬∃yP(x,y)]. But at this point I have two hypothesis that contain negations of existencial quantifier, and I don't know how to use them constructively.
I found some other similar questions, but all the answers given do not say which rules must be applied, and since I'm a beginner I didn't understand them.
logic first-order-logic quantifiers natural-deduction formal-proofs
edited Jan 26 at 18:40
Bram28
63.8k44793
asked Jan 25 at 20:55
Damiano ScevolaDamiano Scevola
182
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add a comment |
$begingroup$
I have been stuck with this problem for a long time, I tried reductio ad absurdum and I got the hypothesys [¬∃x∀y¬P(x,y)], then I try to eliminate the negation of the premise, but I have to prove ∀x∃yP(x,y), and after using the introduction of universal quantifier rule, I go again with reductio ad absurdum, gaining a second hypothesis [¬∃yP(x,y)]. But at this point I have two hypothesis that contain negations of existencial quantifier, and I don't know how to use them constructively.
I found some other similar questions, but all the answers given do not say which rules must be applied, and since I'm a beginner I didn't understand them.
logic first-order-logic quantifiers natural-deduction formal-proofs
edited Jan 26 at 18:40
Bram28
63.8k44793
asked Jan 25 at 20:55
Damiano ScevolaDamiano Scevola
182
$endgroup$
add a comment |
$begingroup$
I have been stuck with this problem for a long time, I tried reductio ad absurdum and I got the hypothesys [¬∃x∀y¬P(x,y)], then I try to eliminate the negation of the premise, but I have to prove ∀x∃yP(x,y), and after using the introduction of universal quantifier rule, I go again with reductio ad absurdum, gaining a second hypothesis [¬∃yP(x,y)]. But at this point I have two hypothesis that contain negations of existencial quantifier, and I don't know how to use them constructively.
I found some other similar questions, but all the answers given do not say which rules must be applied, and since I'm a beginner I didn't understand them.
logic first-order-logic quantifiers natural-deduction formal-proofs
edited Jan 26 at 18:40
Bram28
63.8k44793
asked Jan 25 at 20:55
Damiano ScevolaDamiano Scevola
182
$endgroup$
I have been stuck with this problem for a long time, I tried reductio ad absurdum and I got the hypothesys [¬∃x∀y¬P(x,y)], then I try to eliminate the negation of the premise, but I have to prove ∀x∃yP(x,y), and after using the introduction of universal quantifier rule, I go again with reductio ad absurdum, gaining a second hypothesis [¬∃yP(x,y)]. But at this point I have two hypothesis that contain negations of existencial quantifier, and I don't know how to use them constructively.
I found some other similar questions, but all the answers given do not say which rules must be applied, and since I'm a beginner I didn't understand them.
logic first-order-logic quantifiers natural-deduction formal-proofs
logic first-order-logic quantifiers natural-deduction formal-proofs
edited Jan 26 at 18:40
Bram28
63.8k44793
asked Jan 25 at 20:55
Damiano ScevolaDamiano Scevola
182
edited Jan 26 at 18:40
Bram28
63.8k44793
asked Jan 25 at 20:55
Damiano ScevolaDamiano Scevola
182
edited Jan 26 at 18:40
Bram28
63.8k44793
edited Jan 26 at 18:40
Bram28
63.8k44793
edited Jan 26 at 18:40
Bram28
63.8k44793
63.8k44793
asked Jan 25 at 20:55
Damiano ScevolaDamiano Scevola
182
asked Jan 25 at 20:55
Damiano ScevolaDamiano Scevola
182
asked Jan 25 at 20:55
Damiano ScevolaDamiano Scevola
182
182
add a comment |
add a comment |
1 Answer
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You are doing this exactly right! You just have to derive $forall y neg P(x,y)$ from $neg exists y P(x,y)$
Now, I am not sure how your proof system defines the rule for $forall$ Introduction ... in the system that I use you designate a 'fresh' constant to take the role of the arbitrary object from the domain. So this is what it looks like in my preferred system, called Fitch:
answered Jan 25 at 21:09
Bram28Bram28
63.8k44793
$endgroup$
add a comment |
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$begingroup$
You are doing this exactly right! You just have to derive $forall y neg P(x,y)$ from $neg exists y P(x,y)$
Now, I am not sure how your proof system defines the rule for $forall$ Introduction ... in the system that I use you designate a 'fresh' constant to take the role of the arbitrary object from the domain. So this is what it looks like in my preferred system, called Fitch:
answered Jan 25 at 21:09
Bram28Bram28
63.8k44793
$endgroup$
add a comment |
$begingroup$
You are doing this exactly right! You just have to derive $forall y neg P(x,y)$ from $neg exists y P(x,y)$
Now, I am not sure how your proof system defines the rule for $forall$ Introduction ... in the system that I use you designate a 'fresh' constant to take the role of the arbitrary object from the domain. So this is what it looks like in my preferred system, called Fitch:
answered Jan 25 at 21:09
Bram28Bram28
63.8k44793
$endgroup$
add a comment |
$begingroup$
You are doing this exactly right! You just have to derive $forall y neg P(x,y)$ from $neg exists y P(x,y)$
Now, I am not sure how your proof system defines the rule for $forall$ Introduction ... in the system that I use you designate a 'fresh' constant to take the role of the arbitrary object from the domain. So this is what it looks like in my preferred system, called Fitch:
answered Jan 25 at 21:09
Bram28Bram28
63.8k44793
$endgroup$
You are doing this exactly right! You just have to derive $forall y neg P(x,y)$ from $neg exists y P(x,y)$
Now, I am not sure how your proof system defines the rule for $forall$ Introduction ... in the system that I use you designate a 'fresh' constant to take the role of the arbitrary object from the domain. So this is what it looks like in my preferred system, called Fitch:
answered Jan 25 at 21:09
Bram28Bram28
63.8k44793
edited Jan 25 at 21:18
answered Jan 25 at 21:09
Bram28Bram28
63.8k44793
answered Jan 25 at 21:09
Bram28Bram28
63.8k44793
answered Jan 25 at 21:09
Bram28Bram28
63.8k44793
63.8k44793
add a comment |
add a comment |
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