Theorem csbiebg 3176

Description: Bidirectional conversion between an implicit class substitution hypothesis x = A → B = C and its explicit substitution equivalent. (Contributed by NM, 24-Mar-2013.) (Revised by Mario Carneiro, 11-Dec-2016.)

Hypothesis

Ref	Expression
csbiebg.2	⊢ ℲxC

Assertion

Ref	Expression
csbiebg	⊢ (A ∈ V → (∀x(x = A → B = C) ↔ [A / x]B = C))

Distinct variable group:   x,A

Allowed substitution hints:   B(x)   C(x)   V(x)

Proof of Theorem csbiebg

Dummy variable a is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eqeq2 2362	. . . 4 ⊢ (a = A → (x = a ↔ x = A))
2	1	imbi1d 308	. . 3 ⊢ (a = A → ((x = a → B = C) ↔ (x = A → B = C)))
3	2	albidv 1625	. 2 ⊢ (a = A → (∀x(x = a → B = C) ↔ ∀x(x = A → B = C)))
4		csbeq1 3140	. . 3 ⊢ (a = A → [a / x]B = [A / x]B)
5	4	eqeq1d 2361	. 2 ⊢ (a = A → ([a / x]B = C ↔ [A / x]B = C))
6		vex 2863	. . 3 ⊢ a ∈ V
7		csbiebg.2	. . 3 ⊢ ℲxC
8	6, 7	csbieb 3175	. 2 ⊢ (∀x(x = a → B = C) ↔ [a / x]B = C)
9	3, 5, 8	vtoclbg 2916	1 ⊢ (A ∈ V → (∀x(x = A → B = C) ↔ [A / x]B = C))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176  ∀wal 1540   = wceq 1642   ∈ wcel 1710  Ⅎwnfc 2477  [csb 3137

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-3an 936  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-sbc 3048  df-csb 3138

This theorem is used by: (None)