Natural Deduction Proof - Does this work?
I've been practising but I'm not sure of any means to check my work, can anyone point out if they're any mistakes in this proof?
logic first-order-logic
edited Nov 24 '18 at 5:06
Valentin Montmirail
1,74611641
asked Nov 23 '18 at 1:37
esperskiesperski
162
add a comment |
I've been practising but I'm not sure of any means to check my work, can anyone point out if they're any mistakes in this proof?
logic first-order-logic
edited Nov 24 '18 at 5:06
Valentin Montmirail
1,74611641
asked Nov 23 '18 at 1:37
esperskiesperski
162
add a comment |
I've been practising but I'm not sure of any means to check my work, can anyone point out if they're any mistakes in this proof?
logic first-order-logic
edited Nov 24 '18 at 5:06
Valentin Montmirail
1,74611641
asked Nov 23 '18 at 1:37
esperskiesperski
162
I've been practising but I'm not sure of any means to check my work, can anyone point out if they're any mistakes in this proof?
logic first-order-logic
logic first-order-logic
edited Nov 24 '18 at 5:06
Valentin Montmirail
1,74611641
asked Nov 23 '18 at 1:37
esperskiesperski
162
edited Nov 24 '18 at 5:06
Valentin Montmirail
1,74611641
asked Nov 23 '18 at 1:37
esperskiesperski
162
edited Nov 24 '18 at 5:06
Valentin Montmirail
1,74611641
edited Nov 24 '18 at 5:06
Valentin Montmirail
1,74611641
edited Nov 24 '18 at 5:06
Valentin Montmirail
1,74611641
1,74611641
asked Nov 23 '18 at 1:37
esperskiesperski
162
asked Nov 23 '18 at 1:37
esperskiesperski
162
asked Nov 23 '18 at 1:37
esperskiesperski
162
162
add a comment |
add a comment |
1 Answer
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I have no idea what is “subcomp” in your proof, and how you get 2.
I will not give you the detailed answer, because it looks like homeworks.
The theorem you would like to prove is a conjunction (AND), which means you have to prove both sides: P, and ¬Q.
To prove P, you will have to use reduction to absurdity (from ¬A if you can prove ⊥, then you can prove A), or another of its forms (excluded middle or double negation).
¬Q is easier to prove, just use the ¬ rules.
answered Dec 3 '18 at 15:48
TomTom
4221516
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I have no idea what is “subcomp” in your proof, and how you get 2.
I will not give you the detailed answer, because it looks like homeworks.
The theorem you would like to prove is a conjunction (AND), which means you have to prove both sides: P, and ¬Q.
To prove P, you will have to use reduction to absurdity (from ¬A if you can prove ⊥, then you can prove A), or another of its forms (excluded middle or double negation).
¬Q is easier to prove, just use the ¬ rules.
answered Dec 3 '18 at 15:48
TomTom
4221516
add a comment |
I have no idea what is “subcomp” in your proof, and how you get 2.
I will not give you the detailed answer, because it looks like homeworks.
The theorem you would like to prove is a conjunction (AND), which means you have to prove both sides: P, and ¬Q.
To prove P, you will have to use reduction to absurdity (from ¬A if you can prove ⊥, then you can prove A), or another of its forms (excluded middle or double negation).
¬Q is easier to prove, just use the ¬ rules.
answered Dec 3 '18 at 15:48
TomTom
4221516
add a comment |
I have no idea what is “subcomp” in your proof, and how you get 2.
I will not give you the detailed answer, because it looks like homeworks.
The theorem you would like to prove is a conjunction (AND), which means you have to prove both sides: P, and ¬Q.
To prove P, you will have to use reduction to absurdity (from ¬A if you can prove ⊥, then you can prove A), or another of its forms (excluded middle or double negation).
¬Q is easier to prove, just use the ¬ rules.
answered Dec 3 '18 at 15:48
TomTom
4221516
I have no idea what is “subcomp” in your proof, and how you get 2.
I will not give you the detailed answer, because it looks like homeworks.
The theorem you would like to prove is a conjunction (AND), which means you have to prove both sides: P, and ¬Q.
To prove P, you will have to use reduction to absurdity (from ¬A if you can prove ⊥, then you can prove A), or another of its forms (excluded middle or double negation).
¬Q is easier to prove, just use the ¬ rules.
answered Dec 3 '18 at 15:48
TomTom
4221516
answered Dec 3 '18 at 15:48
TomTom
4221516
answered Dec 3 '18 at 15:48
TomTom
4221516
answered Dec 3 '18 at 15:48
TomTom
4221516
4221516
add a comment |
add a comment |
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