Theorem sbiedvw 2132

Description: Conversion of implicit substitution to explicit substitution (deduction version of sbievw 2131). 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 2128	. . 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 2131	. . 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 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

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

This theorem is used by:  2sbievw  2133  iscatd2  17773  bj-elabd2ALT  37671