Theorem sbiedvw 2131

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbievw 2130). Version of sbied 2532 and sbiedv 2533 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 2127	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ (𝜑 → [𝑦 / 𝑥]𝜓))
2		sbiedvw.1	. . . . . 6 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝜓 ↔ 𝜒))
3	2	expcom 419	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 → (𝜓 ↔ 𝜒)))
4	3	pm5.74d 276	. . . 4 ⊢ (𝑥 = 𝑦 → ((𝜑 → 𝜓) ↔ (𝜑 → 𝜒)))
5	4	sbievw 2130	. . 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 401  [wsb 2098

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-sb 2099

This theorem is used by:  2sbievw  2132  iscatd2  17856  bj-elabd2ALT  37913