Theorem sbcie2g 3778

Description: Conversion of implicit substitution to explicit class substitution. This version of sbcie 3779 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 3740	. 2 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
2		sbcie2g.2	. 2 ⊢ (𝑦 = 𝐴 → (𝜓 ↔ 𝜒))
3		sbsbc 3742	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)
4		sbcie2g.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
5	4	sbievw 2130	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
6	3, 5	bitr3i 280	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
7	1, 2, 6	vtoclbg 3519	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝜑 ↔ 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569  [wsb 2098   ∈ wcel 2144  [wsbc 3738

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  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-sbc 3739

This theorem is used by:  sbcel2gv  3804  csbie2g  3886  brab1  5152  bnj90  35400  bnj124  35548  riotasvd  40107