Theorem sbied 2498

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbie 2497) Usage of this theorem is discouraged because it depends on ax-13 2367. See sbiedw 2305, sbiedvw 2089 for variants using disjoint variables, but requiring fewer axioms. (Contributed by NM, 30-Jun-1994.) (Revised by Mario Carneiro, 4-Oct-2016.) (Proof shortened by Wolf Lammen, 24-Jun-2018.) (New usage is discouraged.)

Hypotheses

Ref	Expression
sbied.1	⊢ Ⅎ𝑥𝜑
sbied.2	⊢ (𝜑 → Ⅎ𝑥𝜒)
sbied.3	⊢ (𝜑 → (𝑥 = 𝑦 → (𝜓 ↔ 𝜒)))

Assertion

Ref	Expression
sbied	⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ 𝜒))

Proof of Theorem sbied

Step	Hyp	Ref	Expression
1		sbied.1	. . . 4 ⊢ Ⅎ𝑥𝜑
2	1	sbrim 2294	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ (𝜑 → [𝑦 / 𝑥]𝜓))
3		sbied.2	. . . . 5 ⊢ (𝜑 → Ⅎ𝑥𝜒)
4	1, 3	nfim1 2188	. . . 4 ⊢ Ⅎ𝑥(𝜑 → 𝜒)
5		sbied.3	. . . . . 6 ⊢ (𝜑 → (𝑥 = 𝑦 → (𝜓 ↔ 𝜒)))
6	5	com12 32	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 → (𝜓 ↔ 𝜒)))
7	6	pm5.74d 273	. . . 4 ⊢ (𝑥 = 𝑦 → ((𝜑 → 𝜓) ↔ (𝜑 → 𝜒)))
8	4, 7	sbie 2497	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ (𝜑 → 𝜒))
9	2, 8	bitr3i 277	. 2 ⊢ ((𝜑 → [𝑦 / 𝑥]𝜓) ↔ (𝜑 → 𝜒))
10	9	pm5.74ri 272	1 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ 𝜒))

Syntax hints:   → wi 4   ↔ wb 205  Ⅎwnf 1778  [wsb 2060

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-10 2130  ax-12 2167  ax-13 2367

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-ex 1775  df-nf 1779  df-sb 2061

This theorem is referenced by:  sbiedv  2499  sbco2  2506  wl-equsb3  37023