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Mirrors > Home > ILE Home > Th. List > impbidd | GIF version |
Description: Deduce an equivalence from two implications. (Contributed by Rodolfo Medina, 12-Oct-2010.) |
Ref | Expression |
---|---|
impbidd.1 | ⊢ (𝜑 → (𝜓 → (𝜒 → 𝜃))) |
impbidd.2 | ⊢ (𝜑 → (𝜓 → (𝜃 → 𝜒))) |
Ref | Expression |
---|---|
impbidd | ⊢ (𝜑 → (𝜓 → (𝜒 ↔ 𝜃))) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | impbidd.1 | . 2 ⊢ (𝜑 → (𝜓 → (𝜒 → 𝜃))) | |
2 | impbidd.2 | . 2 ⊢ (𝜑 → (𝜓 → (𝜃 → 𝜒))) | |
3 | bi3 118 | . 2 ⊢ ((𝜒 → 𝜃) → ((𝜃 → 𝜒) → (𝜒 ↔ 𝜃))) | |
4 | 1, 2, 3 | syl6c 66 | 1 ⊢ (𝜑 → (𝜓 → (𝜒 ↔ 𝜃))) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ↔ wb 104 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia2 106 ax-ia3 107 |
This theorem depends on definitions: df-bi 116 |
This theorem is referenced by: impbid21d 127 pm5.74 178 con1biimdc 863 pclem6 1363 |
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