Theorem sbceq2g 3106

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

Assertion

Ref	Expression
sbceq2g	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ 𝐵 = ⦋𝐴 / 𝑥⦌𝐶))

Distinct variable group:   𝑥,𝐵

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

Proof of Theorem sbceq2g

Step	Hyp	Ref	Expression
1		sbceqg 3100	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3098	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐵 = 𝐵)
3	2	eqeq1d 2205	. 2 ⊢ (𝐴 ∈ 𝑉 → (⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
4	1, 3	bitrd 188	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ 𝐵 = ⦋𝐴 / 𝑥⦌𝐶))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1364   ∈ wcel 2167  [wsbc 2989  ⦋csb 3084

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-v 2765  df-sbc 2990  df-csb 3085

This theorem is referenced by:  csbsng  3683  f1od2  6293