I'm stuck and having trouble with ¬P ∨ Q Prove: P → QHelp with simple deductive proofInvalid arguments with true premises and true conclusionIntroductory Natural Deduction QuestionHow to prove the axiom is wrong?What 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 the tautology ¬(P↔¬P) using Fitch?How do you prove B v A |- A v B?S5 proof of ⊢◻(◻P→◻Q)∨◻(◻Q→◻P)trouble with rules of inference practice problems
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I'm stuck and having trouble with ¬P ∨ Q Prove: P → Q
Help with simple deductive proofInvalid arguments with true premises and true conclusionIntroductory Natural Deduction QuestionHow to prove the axiom is wrong?What 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 the tautology ¬(P↔¬P) using Fitch?How do you prove B v A |- A v B?S5 proof of ⊢◻(◻P→◻Q)∨◻(◻Q→◻P)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
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
3
Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org
– Frank Hubeny
Apr 23 at 1:08
1
Welcome to PSE. The answers hint at how to find a proof. But your question seems to be whether proving P → Q from ¬P ∨ Q is the same as proving ¬P ∨ Q from P → Q to which the answer is "No, this is not the same thing, though the proofs might look (structurally or otherwise) similar".
– Jishin Noben
2 days ago
The two statements should have the same truth values. Would that analysis assist in showing equivalence?
– Mark Andrews
2 days 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
edited 2 days ago
Glorfindel
1851210
1851210
New contributor
asked Apr 23 at 0:53
Hamish DochertyHamish Docherty
102
102
New contributor
New contributor
3
Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org
– Frank Hubeny
Apr 23 at 1:08
1
Welcome to PSE. The answers hint at how to find a proof. But your question seems to be whether proving P → Q from ¬P ∨ Q is the same as proving ¬P ∨ Q from P → Q to which the answer is "No, this is not the same thing, though the proofs might look (structurally or otherwise) similar".
– Jishin Noben
2 days ago
The two statements should have the same truth values. Would that analysis assist in showing equivalence?
– Mark Andrews
2 days ago
add a comment |
3
Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org
– Frank Hubeny
Apr 23 at 1:08
1
Welcome to PSE. The answers hint at how to find a proof. But your question seems to be whether proving P → Q from ¬P ∨ Q is the same as proving ¬P ∨ Q from P → Q to which the answer is "No, this is not the same thing, though the proofs might look (structurally or otherwise) similar".
– Jishin Noben
2 days ago
The two statements should have the same truth values. Would that analysis assist in showing equivalence?
– Mark Andrews
2 days ago
3
3
Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org
– Frank Hubeny
Apr 23 at 1:08
Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org
– Frank Hubeny
Apr 23 at 1:08
1
1
Welcome to PSE. The answers hint at how to find a proof. But your question seems to be whether proving P → Q from ¬P ∨ Q is the same as proving ¬P ∨ Q from P → Q to which the answer is "No, this is not the same thing, though the proofs might look (structurally or otherwise) similar".
– Jishin Noben
2 days ago
Welcome to PSE. The answers hint at how to find a proof. But your question seems to be whether proving P → Q from ¬P ∨ Q is the same as proving ¬P ∨ Q from P → Q to which the answer is "No, this is not the same thing, though the proofs might look (structurally or otherwise) similar".
– Jishin Noben
2 days ago
The two statements should have the same truth values. Would that analysis assist in showing equivalence?
– Mark Andrews
2 days ago
The two statements should have the same truth values. Would that analysis assist in showing equivalence?
– Mark Andrews
2 days ago
add a comment |
3 Answers
3
active
oldest
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 this particular case, the two statements are equivalent:(¬P ∨ Q) ⊢ (P → Q)
and (P → Q) ⊢ (¬P ∨ Q)
are both provably true statements, so (¬P ∨ Q) ≡ (P → Q)
.
But in order to prove that equivalence, we need to prove both directions separately. To see why, consider the case where instead of (¬P ∨ Q)
and (P → Q)
, we have these two statements:
P
P ∨ Q
We can trivially prove that (P ∨ Q)
follows from P
; this is more or less the definition of the addition rule. But P
does not necessarily follow from (P ∨ Q)
, since (¬P ∧ Q)
also satisfies that clause. We can prove it in one direction, but they are not equivalent statements.
The two statements are logically equivalent using the material implication rule. Draw out a truth table and see for yourself that the tables are 100 percent identical.
– Logikal
2 days ago
@Logikal I know they are equivalent. I said they are equivalent. But the OP isn't asking if they're equivalent; they are asking if, having already proven A ⊢ B, you must still prove B ⊢ A in order to establish that an equivalence exists. The answer to that is "yes". The fact that this particular equivalence has already been proven doesn't change that.
– Ray
2 days ago
add a comment |
If one uses a truth table generator one can show that (¬P ∨ Q) ↔ (P → Q). To see this, insert the following input into the Stanford Truth Table Tool: (~P or Q)<=>(P=>Q)
This shows that the two statements are equivalent.
However, the question asks one to prove P → Q given the premise ¬P ∨ Q. Here is how one might do this using a natural deduction proof checker:
See the links below for an explanation of the rules used by this proof.
The proof would not be the same if we wanted to prove ¬P ∨ Q given the premise P → Q. That proof would look like this:
Although the statements are equivalent based on a truth table generator, the proofs from one to the other may be different depending on which is the premise and which the conclusion.
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
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 Apr 23 at 4:47
Graham KempGraham Kemp
1,08919
1,08919
add a comment |
add a comment |
In this particular case, the two statements are equivalent:(¬P ∨ Q) ⊢ (P → Q)
and (P → Q) ⊢ (¬P ∨ Q)
are both provably true statements, so (¬P ∨ Q) ≡ (P → Q)
.
But in order to prove that equivalence, we need to prove both directions separately. To see why, consider the case where instead of (¬P ∨ Q)
and (P → Q)
, we have these two statements:
P
P ∨ Q
We can trivially prove that (P ∨ Q)
follows from P
; this is more or less the definition of the addition rule. But P
does not necessarily follow from (P ∨ Q)
, since (¬P ∧ Q)
also satisfies that clause. We can prove it in one direction, but they are not equivalent statements.
The two statements are logically equivalent using the material implication rule. Draw out a truth table and see for yourself that the tables are 100 percent identical.
– Logikal
2 days ago
@Logikal I know they are equivalent. I said they are equivalent. But the OP isn't asking if they're equivalent; they are asking if, having already proven A ⊢ B, you must still prove B ⊢ A in order to establish that an equivalence exists. The answer to that is "yes". The fact that this particular equivalence has already been proven doesn't change that.
– Ray
2 days ago
add a comment |
In this particular case, the two statements are equivalent:(¬P ∨ Q) ⊢ (P → Q)
and (P → Q) ⊢ (¬P ∨ Q)
are both provably true statements, so (¬P ∨ Q) ≡ (P → Q)
.
But in order to prove that equivalence, we need to prove both directions separately. To see why, consider the case where instead of (¬P ∨ Q)
and (P → Q)
, we have these two statements:
P
P ∨ Q
We can trivially prove that (P ∨ Q)
follows from P
; this is more or less the definition of the addition rule. But P
does not necessarily follow from (P ∨ Q)
, since (¬P ∧ Q)
also satisfies that clause. We can prove it in one direction, but they are not equivalent statements.
The two statements are logically equivalent using the material implication rule. Draw out a truth table and see for yourself that the tables are 100 percent identical.
– Logikal
2 days ago
@Logikal I know they are equivalent. I said they are equivalent. But the OP isn't asking if they're equivalent; they are asking if, having already proven A ⊢ B, you must still prove B ⊢ A in order to establish that an equivalence exists. The answer to that is "yes". The fact that this particular equivalence has already been proven doesn't change that.
– Ray
2 days ago
add a comment |
In this particular case, the two statements are equivalent:(¬P ∨ Q) ⊢ (P → Q)
and (P → Q) ⊢ (¬P ∨ Q)
are both provably true statements, so (¬P ∨ Q) ≡ (P → Q)
.
But in order to prove that equivalence, we need to prove both directions separately. To see why, consider the case where instead of (¬P ∨ Q)
and (P → Q)
, we have these two statements:
P
P ∨ Q
We can trivially prove that (P ∨ Q)
follows from P
; this is more or less the definition of the addition rule. But P
does not necessarily follow from (P ∨ Q)
, since (¬P ∧ Q)
also satisfies that clause. We can prove it in one direction, but they are not equivalent statements.
In this particular case, the two statements are equivalent:(¬P ∨ Q) ⊢ (P → Q)
and (P → Q) ⊢ (¬P ∨ Q)
are both provably true statements, so (¬P ∨ Q) ≡ (P → Q)
.
But in order to prove that equivalence, we need to prove both directions separately. To see why, consider the case where instead of (¬P ∨ Q)
and (P → Q)
, we have these two statements:
P
P ∨ Q
We can trivially prove that (P ∨ Q)
follows from P
; this is more or less the definition of the addition rule. But P
does not necessarily follow from (P ∨ Q)
, since (¬P ∧ Q)
also satisfies that clause. We can prove it in one direction, but they are not equivalent statements.
answered 2 days ago
RayRay
27217
27217
The two statements are logically equivalent using the material implication rule. Draw out a truth table and see for yourself that the tables are 100 percent identical.
– Logikal
2 days ago
@Logikal I know they are equivalent. I said they are equivalent. But the OP isn't asking if they're equivalent; they are asking if, having already proven A ⊢ B, you must still prove B ⊢ A in order to establish that an equivalence exists. The answer to that is "yes". The fact that this particular equivalence has already been proven doesn't change that.
– Ray
2 days ago
add a comment |
The two statements are logically equivalent using the material implication rule. Draw out a truth table and see for yourself that the tables are 100 percent identical.
– Logikal
2 days ago
@Logikal I know they are equivalent. I said they are equivalent. But the OP isn't asking if they're equivalent; they are asking if, having already proven A ⊢ B, you must still prove B ⊢ A in order to establish that an equivalence exists. The answer to that is "yes". The fact that this particular equivalence has already been proven doesn't change that.
– Ray
2 days ago
The two statements are logically equivalent using the material implication rule. Draw out a truth table and see for yourself that the tables are 100 percent identical.
– Logikal
2 days ago
The two statements are logically equivalent using the material implication rule. Draw out a truth table and see for yourself that the tables are 100 percent identical.
– Logikal
2 days ago
@Logikal I know they are equivalent. I said they are equivalent. But the OP isn't asking if they're equivalent; they are asking if, having already proven A ⊢ B, you must still prove B ⊢ A in order to establish that an equivalence exists. The answer to that is "yes". The fact that this particular equivalence has already been proven doesn't change that.
– Ray
2 days ago
@Logikal I know they are equivalent. I said they are equivalent. But the OP isn't asking if they're equivalent; they are asking if, having already proven A ⊢ B, you must still prove B ⊢ A in order to establish that an equivalence exists. The answer to that is "yes". The fact that this particular equivalence has already been proven doesn't change that.
– Ray
2 days ago
add a comment |
If one uses a truth table generator one can show that (¬P ∨ Q) ↔ (P → Q). To see this, insert the following input into the Stanford Truth Table Tool: (~P or Q)<=>(P=>Q)
This shows that the two statements are equivalent.
However, the question asks one to prove P → Q given the premise ¬P ∨ Q. Here is how one might do this using a natural deduction proof checker:
See the links below for an explanation of the rules used by this proof.
The proof would not be the same if we wanted to prove ¬P ∨ Q given the premise P → Q. That proof would look like this:
Although the statements are equivalent based on a truth table generator, the proofs from one to the other may be different depending on which is the premise and which the conclusion.
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/
add a comment |
If one uses a truth table generator one can show that (¬P ∨ Q) ↔ (P → Q). To see this, insert the following input into the Stanford Truth Table Tool: (~P or Q)<=>(P=>Q)
This shows that the two statements are equivalent.
However, the question asks one to prove P → Q given the premise ¬P ∨ Q. Here is how one might do this using a natural deduction proof checker:
See the links below for an explanation of the rules used by this proof.
The proof would not be the same if we wanted to prove ¬P ∨ Q given the premise P → Q. That proof would look like this:
Although the statements are equivalent based on a truth table generator, the proofs from one to the other may be different depending on which is the premise and which the conclusion.
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/
add a comment |
If one uses a truth table generator one can show that (¬P ∨ Q) ↔ (P → Q). To see this, insert the following input into the Stanford Truth Table Tool: (~P or Q)<=>(P=>Q)
This shows that the two statements are equivalent.
However, the question asks one to prove P → Q given the premise ¬P ∨ Q. Here is how one might do this using a natural deduction proof checker:
See the links below for an explanation of the rules used by this proof.
The proof would not be the same if we wanted to prove ¬P ∨ Q given the premise P → Q. That proof would look like this:
Although the statements are equivalent based on a truth table generator, the proofs from one to the other may be different depending on which is the premise and which the conclusion.
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/
If one uses a truth table generator one can show that (¬P ∨ Q) ↔ (P → Q). To see this, insert the following input into the Stanford Truth Table Tool: (~P or Q)<=>(P=>Q)
This shows that the two statements are equivalent.
However, the question asks one to prove P → Q given the premise ¬P ∨ Q. Here is how one might do this using a natural deduction proof checker:
See the links below for an explanation of the rules used by this proof.
The proof would not be the same if we wanted to prove ¬P ∨ Q given the premise P → Q. That proof would look like this:
Although the statements are equivalent based on a truth table generator, the proofs from one to the other may be different depending on which is the premise and which the conclusion.
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/
answered yesterday
Frank HubenyFrank Hubeny
10.8k51559
10.8k51559
add a comment |
add a comment |
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
Apr 23 at 1:08
1
Welcome to PSE. The answers hint at how to find a proof. But your question seems to be whether proving P → Q from ¬P ∨ Q is the same as proving ¬P ∨ Q from P → Q to which the answer is "No, this is not the same thing, though the proofs might look (structurally or otherwise) similar".
– Jishin Noben
2 days ago
The two statements should have the same truth values. Would that analysis assist in showing equivalence?
– Mark Andrews
2 days ago