Theorem sbceq1g 4196

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 4192	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3752	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
3	2	eqeq2d 2827	. 2 ⊢ (𝐴 ∈ 𝑉 → (⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = 𝐶))
4	1, 3	bitrd 270	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = 𝐶))

Syntax hints:   → wi 4   ↔ wb 197   = wceq 1637   ∈ wcel 2157  [wsbc 3644  ⦋csb 3739

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-5 2001  ax-6 2069  ax-7 2105  ax-9 2166  ax-10 2186  ax-11 2202  ax-12 2215  ax-13 2422  ax-ext 2795

This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-tru 1641  df-ex 1860  df-nf 1864  df-sb 2062  df-clab 2804  df-cleq 2810  df-clel 2813  df-nfc 2948  df-v 3404  df-sbc 3645  df-csb 3740

This theorem is referenced by:  telgsums  18599  suppss2f  29776  f1od2  29836  finxpreclem4  33553  frege120  38782