How to Prove P(a) → ∀x(P(x) ∨ ¬(x = a)) using Natural Deduction
How would a formal Fitch proof look like.
I am given P(a) → ∀x(P(x) ∨ ¬(x = a)) to prove using Natural Deduction of predicate logic.
I am confused on how to proceed with the proof.
Please advice me on how to go about with this.
Thanks in advance
logic proof fitch quantification
asked Apr 2 at 23:56
Moey mnmMoey mnm
264
add a comment |
How would a formal Fitch proof look like.
I am given P(a) → ∀x(P(x) ∨ ¬(x = a)) to prove using Natural Deduction of predicate logic.
I am confused on how to proceed with the proof.
Please advice me on how to go about with this.
Thanks in advance
logic proof fitch quantification
asked Apr 2 at 23:56
Moey mnmMoey mnm
264
add a comment |
How would a formal Fitch proof look like.
I am given P(a) → ∀x(P(x) ∨ ¬(x = a)) to prove using Natural Deduction of predicate logic.
I am confused on how to proceed with the proof.
Please advice me on how to go about with this.
Thanks in advance
logic proof fitch quantification
asked Apr 2 at 23:56
Moey mnmMoey mnm
264
How would a formal Fitch proof look like.
I am given P(a) → ∀x(P(x) ∨ ¬(x = a)) to prove using Natural Deduction of predicate logic.
I am confused on how to proceed with the proof.
Please advice me on how to go about with this.
Thanks in advance
logic proof fitch quantification
logic proof fitch quantification
asked Apr 2 at 23:56
Moey mnmMoey mnm
264
asked Apr 2 at 23:56
Moey mnmMoey mnm
264
asked Apr 2 at 23:56
Moey mnmMoey mnm
264
asked Apr 2 at 23:56
Moey mnmMoey mnm
264
asked Apr 2 at 23:56
Moey mnmMoey mnm
264
264
add a comment |
add a comment |
1 Answer
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HINT: I'll sketch the derivation. Since the theorem is a conditional, try using conditional proof/conditional-introduction by assuming P(a) and trying to derive ∀x(P(x) ∨ ¬(x = a)) from it. Here, to derive it, I would try an indirect proof by assuming the negation ¬∀x(P(x) ∨ ¬(x = a)) and trying to derive a contradiction. Use quantifier equivalence rules to get ∃x¬(P(x) ∨ ¬(x = a)).
The next steps will be a little different depending on your list of rules (quantifier rules typically come with restrictions to ensure the rules are sound, and different texts will use different restrictions). Roughly, we can let y be the individual such that ¬(P(y) ∨ ¬(y = a)). Apply De Morgan's law to get ¬P(y) ∧ (y = a). Since y = a, it must be that ¬P(a), contradicting our assumption that P(a). Hence our contradiction completing the indirect proof of ∀x(P(x) ∨ ¬(x = a)).
Hope this helps!
answered Apr 3 at 1:12
AdamAdam
65219
1
Just as confirmation, your suggestion for how to proceed worked using the following proof checker: proofs.openlogicproject.org
– Frank Hubeny
Apr 3 at 2:44
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
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oldest
votes
HINT: I'll sketch the derivation. Since the theorem is a conditional, try using conditional proof/conditional-introduction by assuming P(a) and trying to derive ∀x(P(x) ∨ ¬(x = a)) from it. Here, to derive it, I would try an indirect proof by assuming the negation ¬∀x(P(x) ∨ ¬(x = a)) and trying to derive a contradiction. Use quantifier equivalence rules to get ∃x¬(P(x) ∨ ¬(x = a)).
The next steps will be a little different depending on your list of rules (quantifier rules typically come with restrictions to ensure the rules are sound, and different texts will use different restrictions). Roughly, we can let y be the individual such that ¬(P(y) ∨ ¬(y = a)). Apply De Morgan's law to get ¬P(y) ∧ (y = a). Since y = a, it must be that ¬P(a), contradicting our assumption that P(a). Hence our contradiction completing the indirect proof of ∀x(P(x) ∨ ¬(x = a)).
Hope this helps!
answered Apr 3 at 1:12
AdamAdam
65219
1
Just as confirmation, your suggestion for how to proceed worked using the following proof checker: proofs.openlogicproject.org
– Frank Hubeny
Apr 3 at 2:44
add a comment |
HINT: I'll sketch the derivation. Since the theorem is a conditional, try using conditional proof/conditional-introduction by assuming P(a) and trying to derive ∀x(P(x) ∨ ¬(x = a)) from it. Here, to derive it, I would try an indirect proof by assuming the negation ¬∀x(P(x) ∨ ¬(x = a)) and trying to derive a contradiction. Use quantifier equivalence rules to get ∃x¬(P(x) ∨ ¬(x = a)).
The next steps will be a little different depending on your list of rules (quantifier rules typically come with restrictions to ensure the rules are sound, and different texts will use different restrictions). Roughly, we can let y be the individual such that ¬(P(y) ∨ ¬(y = a)). Apply De Morgan's law to get ¬P(y) ∧ (y = a). Since y = a, it must be that ¬P(a), contradicting our assumption that P(a). Hence our contradiction completing the indirect proof of ∀x(P(x) ∨ ¬(x = a)).
Hope this helps!
answered Apr 3 at 1:12
AdamAdam
65219
1
Just as confirmation, your suggestion for how to proceed worked using the following proof checker: proofs.openlogicproject.org
– Frank Hubeny
Apr 3 at 2:44
add a comment |
HINT: I'll sketch the derivation. Since the theorem is a conditional, try using conditional proof/conditional-introduction by assuming P(a) and trying to derive ∀x(P(x) ∨ ¬(x = a)) from it. Here, to derive it, I would try an indirect proof by assuming the negation ¬∀x(P(x) ∨ ¬(x = a)) and trying to derive a contradiction. Use quantifier equivalence rules to get ∃x¬(P(x) ∨ ¬(x = a)).
The next steps will be a little different depending on your list of rules (quantifier rules typically come with restrictions to ensure the rules are sound, and different texts will use different restrictions). Roughly, we can let y be the individual such that ¬(P(y) ∨ ¬(y = a)). Apply De Morgan's law to get ¬P(y) ∧ (y = a). Since y = a, it must be that ¬P(a), contradicting our assumption that P(a). Hence our contradiction completing the indirect proof of ∀x(P(x) ∨ ¬(x = a)).
Hope this helps!
answered Apr 3 at 1:12
AdamAdam
65219
HINT: I'll sketch the derivation. Since the theorem is a conditional, try using conditional proof/conditional-introduction by assuming P(a) and trying to derive ∀x(P(x) ∨ ¬(x = a)) from it. Here, to derive it, I would try an indirect proof by assuming the negation ¬∀x(P(x) ∨ ¬(x = a)) and trying to derive a contradiction. Use quantifier equivalence rules to get ∃x¬(P(x) ∨ ¬(x = a)).
The next steps will be a little different depending on your list of rules (quantifier rules typically come with restrictions to ensure the rules are sound, and different texts will use different restrictions). Roughly, we can let y be the individual such that ¬(P(y) ∨ ¬(y = a)). Apply De Morgan's law to get ¬P(y) ∧ (y = a). Since y = a, it must be that ¬P(a), contradicting our assumption that P(a). Hence our contradiction completing the indirect proof of ∀x(P(x) ∨ ¬(x = a)).
Hope this helps!
answered Apr 3 at 1:12
AdamAdam
65219
edited Apr 3 at 18:36
answered Apr 3 at 1:12
AdamAdam
65219
answered Apr 3 at 1:12
AdamAdam
65219
answered Apr 3 at 1:12
AdamAdam
65219
65219
1
Just as confirmation, your suggestion for how to proceed worked using the following proof checker: proofs.openlogicproject.org
– Frank Hubeny
Apr 3 at 2:44
add a comment |
1
Just as confirmation, your suggestion for how to proceed worked using the following proof checker: proofs.openlogicproject.org
– Frank Hubeny
Apr 3 at 2:44
1
1
Just as confirmation, your suggestion for how to proceed worked using the following proof checker: proofs.openlogicproject.org
– Frank Hubeny
Apr 3 at 2:44
Just as confirmation, your suggestion for how to proceed worked using the following proof checker: proofs.openlogicproject.org
– Frank Hubeny
Apr 3 at 2:44
add a comment |
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