Theorem nfbidf 1532

Description: An equality theorem for effectively not free. (Contributed by Mario Carneiro, 4-Oct-2016.)

Hypotheses

Ref	Expression
nfbidf.1	⊢ Ⅎ𝑥𝜑
nfbidf.2	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
nfbidf	⊢ (𝜑 → (Ⅎ𝑥𝜓 ↔ Ⅎ𝑥𝜒))

Proof of Theorem nfbidf

Step	Hyp	Ref	Expression
1		nfbidf.1	. . . 4 ⊢ Ⅎ𝑥𝜑
2	1	nfri 1512	. . 3 ⊢ (𝜑 → ∀𝑥𝜑)
3		nfbidf.2	. . . 4 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
4	2, 3	albidh 1473	. . . 4 ⊢ (𝜑 → (∀𝑥𝜓 ↔ ∀𝑥𝜒))
5	3, 4	imbi12d 233	. . 3 ⊢ (𝜑 → ((𝜓 → ∀𝑥𝜓) ↔ (𝜒 → ∀𝑥𝜒)))
6	2, 5	albidh 1473	. 2 ⊢ (𝜑 → (∀𝑥(𝜓 → ∀𝑥𝜓) ↔ ∀𝑥(𝜒 → ∀𝑥𝜒)))
7		df-nf 1454	. 2 ⊢ (Ⅎ𝑥𝜓 ↔ ∀𝑥(𝜓 → ∀𝑥𝜓))
8		df-nf 1454	. 2 ⊢ (Ⅎ𝑥𝜒 ↔ ∀𝑥(𝜒 → ∀𝑥𝜒))
9	6, 7, 8	3bitr4g 222	1 ⊢ (𝜑 → (Ⅎ𝑥𝜓 ↔ Ⅎ𝑥𝜒))

Syntax hints:   → wi 4   ↔ wb 104  ∀wal 1346  Ⅎwnf 1453

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 1440  ax-gen 1442  ax-4 1503

This theorem depends on definitions:  df-bi 116  df-nf 1454

This theorem is referenced by:  dvelimdf  2009  nfcjust  2300  nfceqdf  2311  nfabdw  2331