Theorem sbied 2533

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbie 2532) Usage of this theorem is discouraged because it depends on ax-13 2402. See sbiedw 2347, sbiedvw 2132 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 2338	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ (𝜑 → [𝑦 / 𝑥]𝜓))
3		sbied.2	. . . . 5 ⊢ (𝜑 → Ⅎ𝑥𝜒)
4	1, 3	nfim1 2236	. . . 4 ⊢ Ⅎ𝑥(𝜑 → 𝜒)
5		sbied.3	. . . . . 6 ⊢ (𝜑 → (𝑥 = 𝑦 → (𝜓 ↔ 𝜒)))
6	5	com12 33	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 → (𝜓 ↔ 𝜒)))
7	6	pm5.74d 276	. . . 4 ⊢ (𝑥 = 𝑦 → ((𝜑 → 𝜓) ↔ (𝜑 → 𝜒)))
8	4, 7	sbie 2532	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ (𝜑 → 𝜒))
9	2, 8	bitr3i 280	. 2 ⊢ ((𝜑 → [𝑦 / 𝑥]𝜓) ↔ (𝜑 → 𝜒))
10	9	pm5.74ri 275	1 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  Ⅎwnf 1816  [wsb 2099

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-10 2178  ax-12 2213  ax-13 2402

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817  df-sb 2100

This theorem is used by:  sbiedv  2534  sbco2  2541  wl-equsb3  38468