Theorem sbiedw 2349

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbiev 2347). Version of sbied 2535 with a disjoint variable condition, requiring fewer axioms. (Contributed by NM, 30-Jun-1994.) Avoid ax-13 2404. (Revised by GG, 10-Jan-2024.)

Hypotheses

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

Assertion

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

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥, 𝑦)   𝜓(𝑥, 𝑦)   𝜒(𝑥, 𝑦)

Proof of Theorem sbiedw

Step	Hyp	Ref	Expression
1		sbiedw.1	. . . 4 ⊢ Ⅎ𝑥𝜑
2	1	sbrim 2339	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ (𝜑 → [𝑦 / 𝑥]𝜓))
3		sbiedw.2	. . . . 5 ⊢ (𝜑 → Ⅎ𝑥𝜒)
4	1, 3	nfim1 2235	. . . 4 ⊢ Ⅎ𝑥(𝜑 → 𝜒)
5		sbiedw.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	sbiev 2347	. . 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 1813  [wsb 2096

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-12 2213

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-ex 1810  df-nf 1814  df-sb 2097

This theorem is used by:  csbie2df  4408