Theorem sbceq1g 4371

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 4366	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3870	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
3	2	eqeq2d 2748	. 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 206   = wceq 1542   ∈ wcel 2114  [wsbc 3742  ⦋csb 3851

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-10 2147  ax-11 2163  ax-12 2185  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-nf 1786  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-nfc 2886  df-v 3444  df-sbc 3743  df-csb 3852

This theorem is referenced by:  telgsums  19934  suppss2f  32728  f1od2  32809  finxpreclem4  37649  rspcsbnea  42501  tfsconcatfv  43698  frege120  44339