Theorem eqsbc1 3086

Description: Substitution for the left-hand side in an equality. Class version of eqsb1 2454. (Contributed by Andrew Salmon, 29-Jun-2011.)

Assertion

Ref	Expression
eqsbc1	⊢ (A ∈ V → ([̣A / x]̣x = B ↔ A = B))

Distinct variable group:   x,B

Allowed substitution hints:   A(x)   V(x)

Proof of Theorem eqsbc1

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfsbcq 3049	. 2 ⊢ (y = A → ([̣y / x]̣x = B ↔ [̣A / x]̣x = B))
2		eqeq1 2359	. 2 ⊢ (y = A → (y = B ↔ A = B))
3		sbsbc 3051	. . 3 ⊢ ([y / x]x = B ↔ [̣y / x]̣x = B)
4		eqsb1 2454	. . 3 ⊢ ([y / x]x = B ↔ y = B)
5	3, 4	bitr3i 242	. 2 ⊢ ([̣y / x]̣x = B ↔ y = B)
6	1, 2, 5	vtoclbg 2916	1 ⊢ (A ∈ V → ([̣A / x]̣x = B ↔ A = B))

Syntax hints:   → wi 4   ↔ wb 176   = wceq 1642  [wsb 1648   ∈ wcel 1710  [̣wsbc 3047

This theorem was proved from 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 theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  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

This theorem is referenced by:  sbceqal  3098  eqsbc2  3104