Theorem sbceq1g 3157

Description: Move proper substitution to first argument of an equality. (Contributed by NM, 30-Nov-2005.)

Assertion

Ref	Expression
sbceq1g	⊢ (A ∈ V → ([̣A / x]̣B = C ↔ [A / x]B = C))

Distinct variable group:   x,C

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

Proof of Theorem sbceq1g

Step	Hyp	Ref	Expression
1		sbceqg 3153	. 2 ⊢ (A ∈ V → ([̣A / x]̣B = C ↔ [A / x]B = [A / x]C))
2		csbconstg 3151	. . 3 ⊢ (A ∈ V → [A / x]C = C)
3	2	eqeq2d 2364	. 2 ⊢ (A ∈ V → ([A / x]B = [A / x]C ↔ [A / x]B = C))
4	1, 3	bitrd 244	1 ⊢ (A ∈ V → ([̣A / x]̣B = C ↔ [A / x]B = C))

Syntax hints:   → wi 4   ↔ wb 176   = wceq 1642   ∈ wcel 1710  [̣wsbc 3047  [csb 3137

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  df-csb 3138

This theorem is referenced by:  eqerlem  5961