Prove: $(Arightarrow B),(Arightarrow C)rightarrow B, mapsto_{HPC} B $
0
I'd really like your help proving:
$(Arightarrow B),(Arightarrow C)rightarrow B, mapsto_{HPC} B $
Where $HPC$ is the Hilbert's system proof which contains the following relevant axioms:
- $Arightarrow(B rightarrow A)$
- $(Arightarrow(Brightarrow C)) to ((Arightarrow B)rightarrow(Arightarrow C))$
- $(Arightarrow B)rightarrow ((Arightarrowbar{B})rightarrow bar{A})$
- $bar{bar{A}} rightarrow A$
In addition tried to use these following lemmas: $bar{A} rightarrow (A rightarrow C) $ and $(Arightarrow B)rightarrow (bar{B} rightarrow bar{A})$.
Any suggestions?
logic propositional-calculus proof-theory hilbert-calculus
edited Jul 29 at 12:25
Taroccoesbrocco
5,01261839
asked Mar 21 '13 at 17:14
Jozef
2,82132564
add a comment |
0
I'd really like your help proving:
$(Arightarrow B),(Arightarrow C)rightarrow B, mapsto_{HPC} B $
Where $HPC$ is the Hilbert's system proof which contains the following relevant axioms:
- $Arightarrow(B rightarrow A)$
- $(Arightarrow(Brightarrow C)) to ((Arightarrow B)rightarrow(Arightarrow C))$
- $(Arightarrow B)rightarrow ((Arightarrowbar{B})rightarrow bar{A})$
- $bar{bar{A}} rightarrow A$
In addition tried to use these following lemmas: $bar{A} rightarrow (A rightarrow C) $ and $(Arightarrow B)rightarrow (bar{B} rightarrow bar{A})$.
Any suggestions?
logic propositional-calculus proof-theory hilbert-calculus
edited Jul 29 at 12:25
Taroccoesbrocco
5,01261839
asked Mar 21 '13 at 17:14
Jozef
2,82132564
add a comment |
0
0
0
I'd really like your help proving:
$(Arightarrow B),(Arightarrow C)rightarrow B, mapsto_{HPC} B $
Where $HPC$ is the Hilbert's system proof which contains the following relevant axioms:
- $Arightarrow(B rightarrow A)$
- $(Arightarrow(Brightarrow C)) to ((Arightarrow B)rightarrow(Arightarrow C))$
- $(Arightarrow B)rightarrow ((Arightarrowbar{B})rightarrow bar{A})$
- $bar{bar{A}} rightarrow A$
In addition tried to use these following lemmas: $bar{A} rightarrow (A rightarrow C) $ and $(Arightarrow B)rightarrow (bar{B} rightarrow bar{A})$.
Any suggestions?
logic propositional-calculus proof-theory hilbert-calculus
edited Jul 29 at 12:25
Taroccoesbrocco
5,01261839
asked Mar 21 '13 at 17:14
Jozef
2,82132564
I'd really like your help proving:
$(Arightarrow B),(Arightarrow C)rightarrow B, mapsto_{HPC} B $
Where $HPC$ is the Hilbert's system proof which contains the following relevant axioms:
- $Arightarrow(B rightarrow A)$
- $(Arightarrow(Brightarrow C)) to ((Arightarrow B)rightarrow(Arightarrow C))$
- $(Arightarrow B)rightarrow ((Arightarrowbar{B})rightarrow bar{A})$
- $bar{bar{A}} rightarrow A$
In addition tried to use these following lemmas: $bar{A} rightarrow (A rightarrow C) $ and $(Arightarrow B)rightarrow (bar{B} rightarrow bar{A})$.
Any suggestions?
logic propositional-calculus proof-theory hilbert-calculus
logic propositional-calculus proof-theory hilbert-calculus
edited Jul 29 at 12:25
Taroccoesbrocco
5,01261839
asked Mar 21 '13 at 17:14
Jozef
2,82132564
edited Jul 29 at 12:25
Taroccoesbrocco
5,01261839
asked Mar 21 '13 at 17:14
Jozef
2,82132564
edited Jul 29 at 12:25
Taroccoesbrocco
5,01261839
edited Jul 29 at 12:25
Taroccoesbrocco
5,01261839
edited Jul 29 at 12:25
Taroccoesbrocco
5,01261839
5,01261839
asked Mar 21 '13 at 17:14
Jozef
2,82132564
asked Mar 21 '13 at 17:14
Jozef
2,82132564
asked Mar 21 '13 at 17:14
Jozef
2,82132564
2,82132564
add a comment |
add a comment |
1 Answer
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- $A→B$ (premise)
- $(A→C)→B$ (premise)
- $(A→B)→(neg B→neg A)$ (your second lemma)
- $neg B→ neg A$ (3,1)
- $neg A →(A→C)$ (your first lemma)
- $(neg A →(A→C))→(neg B → (neg A →(A→C)))$ (axiom 1)
- $neg B → (neg A →(A→C))$ (6,5)
- $(neg B → (neg A →(A→C))) → ((neg B → neg A)→(neg B → (A→C)))$ (axiom 2)
- $((neg B → neg A)→(neg B → (A→C)))$ (8,7)
- $neg B → (A→C)$ (9,4)
- $((A→C)→B)→(neg B → neg(A→C))$ (your second lemma)
- $neg B → neg(A→C)$ (11, 2)
- $(neg B → (A→C))→((neg B → neg(A→C))→neg neg B)$ (axiom 3)
- $(neg B → neg(A→C))→neg neg B$ (13, 10)
- $neg neg B$ (14, 12)
- $neg neg B → B$ (axiom 4)
- $B$ (16, 15)
answered Mar 23 '13 at 17:04
Cian
411215
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
- $A→B$ (premise)
- $(A→C)→B$ (premise)
- $(A→B)→(neg B→neg A)$ (your second lemma)
- $neg B→ neg A$ (3,1)
- $neg A →(A→C)$ (your first lemma)
- $(neg A →(A→C))→(neg B → (neg A →(A→C)))$ (axiom 1)
- $neg B → (neg A →(A→C))$ (6,5)
- $(neg B → (neg A →(A→C))) → ((neg B → neg A)→(neg B → (A→C)))$ (axiom 2)
- $((neg B → neg A)→(neg B → (A→C)))$ (8,7)
- $neg B → (A→C)$ (9,4)
- $((A→C)→B)→(neg B → neg(A→C))$ (your second lemma)
- $neg B → neg(A→C)$ (11, 2)
- $(neg B → (A→C))→((neg B → neg(A→C))→neg neg B)$ (axiom 3)
- $(neg B → neg(A→C))→neg neg B$ (13, 10)
- $neg neg B$ (14, 12)
- $neg neg B → B$ (axiom 4)
- $B$ (16, 15)
answered Mar 23 '13 at 17:04
Cian
411215
add a comment |
4
- $A→B$ (premise)
- $(A→C)→B$ (premise)
- $(A→B)→(neg B→neg A)$ (your second lemma)
- $neg B→ neg A$ (3,1)
- $neg A →(A→C)$ (your first lemma)
- $(neg A →(A→C))→(neg B → (neg A →(A→C)))$ (axiom 1)
- $neg B → (neg A →(A→C))$ (6,5)
- $(neg B → (neg A →(A→C))) → ((neg B → neg A)→(neg B → (A→C)))$ (axiom 2)
- $((neg B → neg A)→(neg B → (A→C)))$ (8,7)
- $neg B → (A→C)$ (9,4)
- $((A→C)→B)→(neg B → neg(A→C))$ (your second lemma)
- $neg B → neg(A→C)$ (11, 2)
- $(neg B → (A→C))→((neg B → neg(A→C))→neg neg B)$ (axiom 3)
- $(neg B → neg(A→C))→neg neg B$ (13, 10)
- $neg neg B$ (14, 12)
- $neg neg B → B$ (axiom 4)
- $B$ (16, 15)
answered Mar 23 '13 at 17:04
Cian
411215
add a comment |
4
4
4
- $A→B$ (premise)
- $(A→C)→B$ (premise)
- $(A→B)→(neg B→neg A)$ (your second lemma)
- $neg B→ neg A$ (3,1)
- $neg A →(A→C)$ (your first lemma)
- $(neg A →(A→C))→(neg B → (neg A →(A→C)))$ (axiom 1)
- $neg B → (neg A →(A→C))$ (6,5)
- $(neg B → (neg A →(A→C))) → ((neg B → neg A)→(neg B → (A→C)))$ (axiom 2)
- $((neg B → neg A)→(neg B → (A→C)))$ (8,7)
- $neg B → (A→C)$ (9,4)
- $((A→C)→B)→(neg B → neg(A→C))$ (your second lemma)
- $neg B → neg(A→C)$ (11, 2)
- $(neg B → (A→C))→((neg B → neg(A→C))→neg neg B)$ (axiom 3)
- $(neg B → neg(A→C))→neg neg B$ (13, 10)
- $neg neg B$ (14, 12)
- $neg neg B → B$ (axiom 4)
- $B$ (16, 15)
answered Mar 23 '13 at 17:04
Cian
411215
- $A→B$ (premise)
- $(A→C)→B$ (premise)
- $(A→B)→(neg B→neg A)$ (your second lemma)
- $neg B→ neg A$ (3,1)
- $neg A →(A→C)$ (your first lemma)
- $(neg A →(A→C))→(neg B → (neg A →(A→C)))$ (axiom 1)
- $neg B → (neg A →(A→C))$ (6,5)
- $(neg B → (neg A →(A→C))) → ((neg B → neg A)→(neg B → (A→C)))$ (axiom 2)
- $((neg B → neg A)→(neg B → (A→C)))$ (8,7)
- $neg B → (A→C)$ (9,4)
- $((A→C)→B)→(neg B → neg(A→C))$ (your second lemma)
- $neg B → neg(A→C)$ (11, 2)
- $(neg B → (A→C))→((neg B → neg(A→C))→neg neg B)$ (axiom 3)
- $(neg B → neg(A→C))→neg neg B$ (13, 10)
- $neg neg B$ (14, 12)
- $neg neg B → B$ (axiom 4)
- $B$ (16, 15)
answered Mar 23 '13 at 17:04
Cian
411215
answered Mar 23 '13 at 17:04
Cian
411215
answered Mar 23 '13 at 17:04
Cian
411215
answered Mar 23 '13 at 17:04
Cian
411215
411215
add a comment |
add a comment |
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