How to Prove P(a) → ∀x(P(x) ∨ ¬(x = a)) using Natural DeductionUsing natural deduction rules give a...
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How to Prove P(a) → ∀x(P(x) ∨ ¬(x = a)) using Natural Deduction
Using natural deduction rules give a formal proofIntroductory Natural Deduction QuestionProve A ∨ D from A ∨ (B ∧ C) and (¬ B ∨ ¬ C) ∨ D ( LPL Q6.26) without using --> or material implicationGiven P ∨ ¬ P prove (P → Q) → ((¬ P → Q) → Q) by natural deductionHow to prove ¬(p→q) ⊢ p &¬qDoes anyone have a proof checker they prefer using for modal logic?How do you prove law of excluded middle using tertium non datur?How to prove : (( P → Q ) ∨ ( Q → R )) by natural deductionHow to prove ‘∃xP(x)’ from ‘¬∀x(P(x)→Q(x))’How would i go about using natural deduction to prove this argument is valid?
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 2 hours ago
Moey mnmMoey mnm
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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 2 hours ago
Moey mnmMoey mnm
16
New contributor
Moey mnm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 hours ago
Moey mnmMoey mnm
16
New contributor
Moey mnm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 hours ago
Moey mnmMoey mnm
16
New contributor
Moey mnm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Moey mnmMoey mnm
16
New contributor
Moey mnm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Moey mnmMoey mnm
16
New contributor
Moey mnm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Moey mnmMoey mnm
16
asked 2 hours ago
Moey mnmMoey mnm
16
16
New contributor
Moey mnm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Moey mnm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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 stand for the particular 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 1 hour ago
AdamAdam
4358
add a comment |
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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 stand for the particular 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 1 hour ago
AdamAdam
4358
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 stand for the particular 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 1 hour ago
AdamAdam
4358
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 stand for the particular 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 1 hour ago
AdamAdam
4358
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 stand for the particular 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 1 hour ago
AdamAdam
4358
answered 1 hour ago
AdamAdam
4358
answered 1 hour ago
AdamAdam
4358
answered 1 hour ago
AdamAdam
4358
4358
add a comment |
add a comment |
Moey mnm is a new contributor. Be nice, and check out our Code of Conduct.
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