Theorem nfbidf 1432

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 1412	. . 3 ⊢ (𝜑 → ∀𝑥𝜑)
3		nfbidf.2	. . . 4 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
4	2, 3	albidh 1369	. . . 4 ⊢ (𝜑 → (∀𝑥𝜓 ↔ ∀𝑥𝜒))
5	3, 4	imbi12d 223	. . 3 ⊢ (𝜑 → ((𝜓 → ∀𝑥𝜓) ↔ (𝜒 → ∀𝑥𝜒)))
6	2, 5	albidh 1369	. 2 ⊢ (𝜑 → (∀𝑥(𝜓 → ∀𝑥𝜓) ↔ ∀𝑥(𝜒 → ∀𝑥𝜒)))
7		df-nf 1350	. 2 ⊢ (Ⅎ𝑥𝜓 ↔ ∀𝑥(𝜓 → ∀𝑥𝜓))
8		df-nf 1350	. 2 ⊢ (Ⅎ𝑥𝜒 ↔ ∀𝑥(𝜒 → ∀𝑥𝜒))
9	6, 7, 8	3bitr4g 212	1 ⊢ (𝜑 → (Ⅎ𝑥𝜓 ↔ Ⅎ𝑥𝜒))

Syntax hints:   → wi 4   ↔ wb 98  ∀wal 1241  Ⅎwnf 1349

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338  ax-4 1400

This theorem depends on definitions:  df-bi 110  df-nf 1350

This theorem is referenced by:  dvelimdf  1892  nfcjust  2166  nfceqdf  2177