Theorem eqsbc1 2994

Description: Substitution for the left-hand side in an equality. Class version of eqsb1 2274. (Contributed by Andrew Salmon, 29-Jun-2011.)

Assertion

Ref	Expression
eqsbc1	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵 ↔ 𝐴 = 𝐵))

Distinct variable group:   𝑥,𝐵

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

Proof of Theorem eqsbc1

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfsbcq 2957	. 2 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ [𝐴 / 𝑥]𝑥 = 𝐵))
2		eqeq1 2177	. 2 ⊢ (𝑦 = 𝐴 → (𝑦 = 𝐵 ↔ 𝐴 = 𝐵))
3		sbsbc 2959	. . 3 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ [𝑦 / 𝑥]𝑥 = 𝐵)
4		eqsb1 2274	. . 3 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ 𝑦 = 𝐵)
5	3, 4	bitr3i 185	. 2 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ 𝑦 = 𝐵)
6	1, 2, 5	vtoclbg 2791	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵 ↔ 𝐴 = 𝐵))

Syntax hints:   → wi 4   ↔ wb 104   = wceq 1348  [wsb 1755   ∈ wcel 2141  [wsbc 2955

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-sbc 2956

This theorem is referenced by:  sbceqal  3010  eqsbc2  3015