Theorem sbiedvw 2129

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbievw 2127). Version of sbied 2534 and sbiedv 2535 with more disjoint variable conditions, requiring fewer axioms. (Contributed by NM, 30-Jun-1994.) (Revised by GG, 29-Jan-2024.)

Hypothesis

Ref	Expression
sbiedvw.1	⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝜓 ↔ 𝜒))

Assertion

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

Distinct variable groups:   𝜑,𝑥   𝜒,𝑥   𝑥,𝑦

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

Proof of Theorem sbiedvw

Step	Hyp	Ref	Expression
1		sbrimvw 2124	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ (𝜑 → [𝑦 / 𝑥]𝜓))
2		sbiedvw.1	. . . . . 6 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝜓 ↔ 𝜒))
3	2	expcom 418	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 → (𝜓 ↔ 𝜒)))
4	3	pm5.74d 276	. . . 4 ⊢ (𝑥 = 𝑦 → ((𝜑 → 𝜓) ↔ (𝜑 → 𝜒)))
5	4	sbievw 2127	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ (𝜑 → 𝜒))
6	1, 5	bitr3i 280	. 2 ⊢ ((𝜑 → [𝑦 / 𝑥]𝜓) ↔ (𝜑 → 𝜒))
7	6	pm5.74ri 275	1 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 400  [wsb 2095

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037

This proof depends on definitions:  df-bi 210  df-an 401  df-ex 1809  df-sb 2096

This theorem is used by:  2sbievw  2130  iscatd2  17743  bj-elabd2ALT  37589