Theorem bisym 225

Description: Express symmetries of theorems in terms of biconditionals. (Contributed by Wolf Lammen, 14-May-2013.)

Assertion

Ref	Expression
bisym	⊢ (((𝜑 → 𝜓) → (𝜒 → 𝜃)) → (((𝜓 → 𝜑) → (𝜃 → 𝜒)) → ((𝜑 ↔ 𝜓) → (𝜒 ↔ 𝜃))))

Proof of Theorem bisym

Step	Hyp	Ref	Expression
1		bi3 119	. 2 ⊢ ((𝜒 → 𝜃) → ((𝜃 → 𝜒) → (𝜒 ↔ 𝜃)))
2	1	bi3ant 224	1 ⊢ (((𝜑 → 𝜓) → (𝜒 → 𝜃)) → (((𝜓 → 𝜑) → (𝜃 → 𝜒)) → ((𝜑 ↔ 𝜓) → (𝜒 ↔ 𝜃))))

This proof depends on syntax axioms:   → wi 4   ↔ wb 105

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This proof depends on definitions:  df-bi 117

This theorem is used by: (None)