Im stuck and having trouble with ¬P ∨ Q Prove: P → Q Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Help with simple deductive proofInvalid arguments with true premises and true conclusionWhat are the important effects of studying logic?If F is a sufficient condition for G, is lacking G a sufficient condition for lacking F?How to prove (P ∧ ¬Q) ↔ ¬(P → Q)Prove (¬P ∨ Q) ↔ (P → Q)How to prove the tautology ¬(P↔¬P) using Fitch?How do you prove B v A |- A v B?I have trouble understanding this fallacy: “If A, then B. Therefore if not-B, then not-A.”trouble with rules of inference practice problems
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Im stuck and having trouble with ¬P ∨ Q Prove: P → Q
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Help with simple deductive proofInvalid arguments with true premises and true conclusionWhat are the important effects of studying logic?If F is a sufficient condition for G, is lacking G a sufficient condition for lacking F?How to prove (P ∧ ¬Q) ↔ ¬(P → Q)Prove (¬P ∨ Q) ↔ (P → Q)How to prove the tautology ¬(P↔¬P) using Fitch?How do you prove B v A |- A v B?I have trouble understanding this fallacy: “If A, then B. Therefore if not-B, then not-A.”trouble with rules of inference practice problems
I am having trouble with this problem as I have just started doing logic. Is this the same as P → Q Prove: ¬P ∨ Q?
logic
New contributor
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I am having trouble with this problem as I have just started doing logic. Is this the same as P → Q Prove: ¬P ∨ Q?
logic
New contributor
2
Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org
– Frank Hubeny
4 hours ago
add a comment |
I am having trouble with this problem as I have just started doing logic. Is this the same as P → Q Prove: ¬P ∨ Q?
logic
New contributor
I am having trouble with this problem as I have just started doing logic. Is this the same as P → Q Prove: ¬P ∨ Q?
logic
logic
New contributor
New contributor
New contributor
asked 5 hours ago
Hamish DochertyHamish Docherty
111
111
New contributor
New contributor
2
Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org
– Frank Hubeny
4 hours ago
add a comment |
2
Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org
– Frank Hubeny
4 hours ago
2
2
Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org
– Frank Hubeny
4 hours ago
Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org
– Frank Hubeny
4 hours ago
add a comment |
1 Answer
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In a natural deduction system (if that is what you are using) to prove a conditional, such as is P → Q, you must use a Conditional
Proof.
This takes the form of assuming the antecedent (that is P) aiming to derive the consequent (that is Q) through valid inferences (also using the premises; that is ¬P ∨ Q). Then discharging the assumption allow the deduction of the conditional (that is P → Q).
Now to prove Q from an assumption of P and the premise of ¬P ∨ Q, either use Disjunctive Syllogism, or a Proof by Cases.
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1 Answer
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1 Answer
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active
oldest
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oldest
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votes
In a natural deduction system (if that is what you are using) to prove a conditional, such as is P → Q, you must use a Conditional
Proof.
This takes the form of assuming the antecedent (that is P) aiming to derive the consequent (that is Q) through valid inferences (also using the premises; that is ¬P ∨ Q). Then discharging the assumption allow the deduction of the conditional (that is P → Q).
Now to prove Q from an assumption of P and the premise of ¬P ∨ Q, either use Disjunctive Syllogism, or a Proof by Cases.
add a comment |
In a natural deduction system (if that is what you are using) to prove a conditional, such as is P → Q, you must use a Conditional
Proof.
This takes the form of assuming the antecedent (that is P) aiming to derive the consequent (that is Q) through valid inferences (also using the premises; that is ¬P ∨ Q). Then discharging the assumption allow the deduction of the conditional (that is P → Q).
Now to prove Q from an assumption of P and the premise of ¬P ∨ Q, either use Disjunctive Syllogism, or a Proof by Cases.
add a comment |
In a natural deduction system (if that is what you are using) to prove a conditional, such as is P → Q, you must use a Conditional
Proof.
This takes the form of assuming the antecedent (that is P) aiming to derive the consequent (that is Q) through valid inferences (also using the premises; that is ¬P ∨ Q). Then discharging the assumption allow the deduction of the conditional (that is P → Q).
Now to prove Q from an assumption of P and the premise of ¬P ∨ Q, either use Disjunctive Syllogism, or a Proof by Cases.
In a natural deduction system (if that is what you are using) to prove a conditional, such as is P → Q, you must use a Conditional
Proof.
This takes the form of assuming the antecedent (that is P) aiming to derive the consequent (that is Q) through valid inferences (also using the premises; that is ¬P ∨ Q). Then discharging the assumption allow the deduction of the conditional (that is P → Q).
Now to prove Q from an assumption of P and the premise of ¬P ∨ Q, either use Disjunctive Syllogism, or a Proof by Cases.
answered 1 hour ago
Graham KempGraham Kemp
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Hamish Docherty is a new contributor. Be nice, and check out our Code of Conduct.
Hamish Docherty is a new contributor. Be nice, and check out our Code of Conduct.
Hamish Docherty is a new contributor. Be nice, and check out our Code of Conduct.
Hamish Docherty is a new contributor. Be nice, and check out our Code of Conduct.
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Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org
– Frank Hubeny
4 hours ago