Theorem eqsbc1 3042

Description: Substitution for the left-hand side in an equality. Class version of eqsb1 2310. (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 3004	. 2 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ [𝐴 / 𝑥]𝑥 = 𝐵))
2		eqeq1 2213	. 2 ⊢ (𝑦 = 𝐴 → (𝑦 = 𝐵 ↔ 𝐴 = 𝐵))
3		sbsbc 3006	. . 3 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ [𝑦 / 𝑥]𝑥 = 𝐵)
4		eqsb1 2310	. . 3 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ 𝑦 = 𝐵)
5	3, 4	bitr3i 186	. 2 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ 𝑦 = 𝐵)
6	1, 2, 5	vtoclbg 2836	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵 ↔ 𝐴 = 𝐵))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1373  [wsb 1786   ∈ wcel 2177  [wsbc 3002

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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-v 2775  df-sbc 3003

This theorem is referenced by:  sbceqal  3058  eqsbc2  3063