Theorem ibibr 245

Description: Implication in terms of implication and biconditional. (Contributed by NM, 29-Apr-2005.) (Proof shortened by Wolf Lammen, 21-Dec-2013.)

Assertion

Ref	Expression
ibibr	⊢ ((𝜑 → 𝜓) ↔ (𝜑 → (𝜓 ↔ 𝜑)))

Proof of Theorem ibibr

Step	Hyp	Ref	Expression
1		pm5.501 243	. . 3 ⊢ (𝜑 → (𝜓 ↔ (𝜑 ↔ 𝜓)))
2		bicom 139	. . 3 ⊢ ((𝜑 ↔ 𝜓) ↔ (𝜓 ↔ 𝜑))
3	1, 2	syl6bb 195	. 2 ⊢ (𝜑 → (𝜓 ↔ (𝜓 ↔ 𝜑)))
4	3	pm5.74i 179	1 ⊢ ((𝜑 → 𝜓) ↔ (𝜑 → (𝜓 ↔ 𝜑)))

Syntax hints:   → wi 4   ↔ wb 104

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107

This theorem depends on definitions:  df-bi 116

This theorem is referenced by:  tbt  246  oibabs  703  rabxfrd  4390