Prove: $(Arightarrow B),(Arightarrow C)rightarrow B, mapsto

Prove: $(Arightarrow B),(Arightarrow C)rightarrow B, mapsto_{HPC} B $

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?

edited Jul 29 '18 at 12:25

Taroccoesbrocco

5,42271839

asked Mar 21 '13 at 17:14

Jozef

2,84132564

add a comment |

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?

edited Jul 29 '18 at 12:25

Taroccoesbrocco

5,42271839

asked Mar 21 '13 at 17:14

Jozef

2,84132564

add a comment |

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?

edited Jul 29 '18 at 12:25

Taroccoesbrocco

5,42271839

asked Mar 21 '13 at 17:14

Jozef

2,84132564

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 '18 at 12:25

Taroccoesbrocco

5,42271839

asked Mar 21 '13 at 17:14

Jozef

2,84132564

edited Jul 29 '18 at 12:25

Taroccoesbrocco

5,42271839

asked Mar 21 '13 at 17:14

Jozef

2,84132564

edited Jul 29 '18 at 12:25

Taroccoesbrocco

5,42271839

edited Jul 29 '18 at 12:25

Taroccoesbrocco

5,42271839

edited Jul 29 '18 at 12:25

Taroccoesbrocco

5,42271839

asked Mar 21 '13 at 17:14

Jozef

2,84132564

asked Mar 21 '13 at 17:14

Jozef

2,84132564

asked Mar 21 '13 at 17:14

Jozef

2,84132564

add a comment |

1 Answer
1

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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

$A→B$ (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 |

$A→B$ (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 |

$A→B$ (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

add a comment |

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