Theorem eqsbc1 3072

Description: Substitution for the left-hand side in an equality. Class version of eqsb1 2335. (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 3034	. 2 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ [𝐴 / 𝑥]𝑥 = 𝐵))
2		eqeq1 2238	. 2 ⊢ (𝑦 = 𝐴 → (𝑦 = 𝐵 ↔ 𝐴 = 𝐵))
3		sbsbc 3036	. . 3 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ [𝑦 / 𝑥]𝑥 = 𝐵)
4		eqsb1 2335	. . 3 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ 𝑦 = 𝐵)
5	3, 4	bitr3i 186	. 2 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ 𝑦 = 𝐵)
6	1, 2, 5	vtoclbg 2866	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵 ↔ 𝐴 = 𝐵))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1398  [wsb 1810   ∈ wcel 2202  [wsbc 3032

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 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-v 2805  df-sbc 3033

This theorem is referenced by:  sbceqal  3088  eqsbc2  3093