Theorem sbceq1g 4440

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 4435	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3940	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
3	2	eqeq2d 2751	. 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 1537   ∈ wcel 2108  [wsbc 3804  ⦋csb 3921

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2158  ax-12 2178  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-tru 1540  df-ex 1778  df-nf 1782  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-nfc 2895  df-v 3490  df-sbc 3805  df-csb 3922

This theorem is referenced by:  telgsums  20035  suppss2f  32657  f1od2  32735  finxpreclem4  37360  rspcsbnea  42088  tfsconcatfv  43303  frege120  43945