Theorem sbcie2g 3786

Description: Conversion of implicit substitution to explicit class substitution. This version of sbcie 3787 avoids a disjointness condition on 𝑥, 𝐴 by substituting twice. (Contributed by Mario Carneiro, 15-Oct-2016.)

Hypotheses

Ref	Expression
sbcie2g.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
sbcie2g.2	⊢ (𝑦 = 𝐴 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
sbcie2g	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝜑 ↔ 𝜒))

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

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

Proof of Theorem sbcie2g

Step	Hyp	Ref	Expression
1		dfsbcq 3748	. 2 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
2		sbcie2g.2	. 2 ⊢ (𝑦 = 𝐴 → (𝜓 ↔ 𝜒))
3		sbsbc 3750	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)
4		sbcie2g.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
5	4	sbievw 2131	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
6	3, 5	bitr3i 280	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
7	1, 2, 6	vtoclbg 3526	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝜑 ↔ 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570  [wsb 2099   ∈ wcel 2146  [wsbc 3746

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-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-sbc 3747

This theorem is used by:  sbcel2gv  3812  csbie2g  3894  brab1  5161  bnj90  35135  bnj124  35283  riotasvd  39763