Theorem eqsbc2 3061

Description: Substitution for the right-hand side in an equality. (Contributed by Alan Sare, 24-Oct-2011.) (Proof shortened by JJ, 7-Jul-2021.)

Assertion

Ref	Expression
eqsbc2	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝑥 ↔ 𝐵 = 𝐴))

Distinct variable group:   𝑥,𝐵

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

Proof of Theorem eqsbc2

Step	Hyp	Ref	Expression
1		eqsbc1 3040	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵 ↔ 𝐴 = 𝐵))
2		eqcom 2208	. . 3 ⊢ (𝐵 = 𝑥 ↔ 𝑥 = 𝐵)
3	2	sbcbii 3060	. 2 ⊢ ([𝐴 / 𝑥]𝐵 = 𝑥 ↔ [𝐴 / 𝑥]𝑥 = 𝐵)
4		eqcom 2208	. 2 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
5	1, 3, 4	3bitr4g 223	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝑥 ↔ 𝐵 = 𝐴))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1373   ∈ wcel 2177  [wsbc 3000

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 3001

This theorem is referenced by: (None)