Theorem sbceq1g 4409

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 4404	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3907	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
3	2	eqeq2d 2737	. 2 ⊢ (𝐴 ∈ 𝑉 → (⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = 𝐶))
4	1, 3	bitrd 279	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = 𝐶))

Syntax hints:   → wi 4   ↔ wb 205   = wceq 1533   ∈ wcel 2098  [wsbc 3772  ⦋csb 3888

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-10 2129  ax-11 2146  ax-12 2163  ax-ext 2697

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1536  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2704  df-cleq 2718  df-clel 2804  df-nfc 2879  df-v 3470  df-sbc 3773  df-csb 3889

This theorem is referenced by:  telgsums  19911  suppss2f  32368  f1od2  32451  finxpreclem4  36782  tfsconcatfv  42648  frege120  43291