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Mirrors > Home > ILE Home > Th. List > pm5.21nd | GIF version |
Description: Eliminate an antecedent implied by each side of a biconditional. (Contributed by NM, 20-Nov-2005.) (Proof shortened by Wolf Lammen, 4-Nov-2013.) |
Ref | Expression |
---|---|
pm5.21nd.1 | ⊢ ((𝜑 ∧ 𝜓) → 𝜃) |
pm5.21nd.2 | ⊢ ((𝜑 ∧ 𝜒) → 𝜃) |
pm5.21nd.3 | ⊢ (𝜃 → (𝜓 ↔ 𝜒)) |
Ref | Expression |
---|---|
pm5.21nd | ⊢ (𝜑 → (𝜓 ↔ 𝜒)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | pm5.21nd.1 | . . 3 ⊢ ((𝜑 ∧ 𝜓) → 𝜃) | |
2 | 1 | ex 115 | . 2 ⊢ (𝜑 → (𝜓 → 𝜃)) |
3 | pm5.21nd.2 | . . 3 ⊢ ((𝜑 ∧ 𝜒) → 𝜃) | |
4 | 3 | ex 115 | . 2 ⊢ (𝜑 → (𝜒 → 𝜃)) |
5 | pm5.21nd.3 | . . 3 ⊢ (𝜃 → (𝜓 ↔ 𝜒)) | |
6 | 5 | a1i 9 | . 2 ⊢ (𝜑 → (𝜃 → (𝜓 ↔ 𝜒))) |
7 | 2, 4, 6 | pm5.21ndd 705 | 1 ⊢ (𝜑 → (𝜓 ↔ 𝜒)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ∧ wa 104 ↔ wb 105 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 |
This theorem depends on definitions: df-bi 117 |
This theorem is referenced by: ideqg 4777 fvelimab 5571 releldm2 6183 relelec 6572 fzrev3 10082 elfzp12 10094 eqgval 13013 eltg 13423 eltg2 13424 cncnp2m 13602 |
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