Theorem bisym 281

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 179	. 2 ⊢ ((χ → θ) → ((θ → χ) → (χ ↔ θ)))
2	1	bi3ant 280	1 ⊢ (((φ → ψ) → (χ → θ)) → (((ψ → φ) → (θ → χ)) → ((φ ↔ ψ) → (χ ↔ θ))))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177

This theorem is used by: (None)