Theorem hbbi 1528

Description: If 𝑥 is not free in 𝜑 and 𝜓, it is not free in (𝜑 ↔ 𝜓). (Contributed by NM, 5-Aug-1993.)

Hypotheses

Ref	Expression
hb.1	⊢ (𝜑 → ∀𝑥𝜑)
hb.2	⊢ (𝜓 → ∀𝑥𝜓)

Assertion

Ref	Expression
hbbi	⊢ ((𝜑 ↔ 𝜓) → ∀𝑥(𝜑 ↔ 𝜓))

Proof of Theorem hbbi

Step	Hyp	Ref	Expression
1		dfbi2 386	. 2 ⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)))
2		hb.1	. . . 4 ⊢ (𝜑 → ∀𝑥𝜑)
3		hb.2	. . . 4 ⊢ (𝜓 → ∀𝑥𝜓)
4	2, 3	hbim 1525	. . 3 ⊢ ((𝜑 → 𝜓) → ∀𝑥(𝜑 → 𝜓))
5	3, 2	hbim 1525	. . 3 ⊢ ((𝜓 → 𝜑) → ∀𝑥(𝜓 → 𝜑))
6	4, 5	hban 1527	. 2 ⊢ (((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) → ∀𝑥((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)))
7	1, 6	hbxfrbi 1449	1 ⊢ ((𝜑 ↔ 𝜓) → ∀𝑥(𝜑 ↔ 𝜓))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104  ∀wal 1330

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1424  ax-gen 1426  ax-4 1488  ax-i5r 1516

This theorem depends on definitions:  df-bi 116

This theorem is referenced by:  euf  2005  sb8euh  2023