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Metamath Proof Explorer

Theorem eqsbc1

Description: Substitution for the left-hand side in an equality. Class version of eqsb1 . (Contributed by Andrew Salmon, 29-Jun-2011) Avoid ax-13 . (Revised by Wolf Lammen, 29-Apr-2023)

Ref Expression
Assertion eqsbc1 ⊢ A ∈ V → [˙A / x]˙ x = B ↔ A = B

Proof

Step Hyp Ref Expression
1 dfsbcq ⊢ y = A → [˙y / x]˙ x = B ↔ [˙A / x]˙ x = B
2 eqeq1 ⊢ y = A → y = B ↔ A = B
3 sbsbc ⊢ y x x = B ↔ [˙y / x]˙ x = B
4 eqsb1 ⊢ y x x = B ↔ y = B
5 3 4 bitr3i ⊢ [˙y / x]˙ x = B ↔ y = B
6 1 2 5 vtoclbg ⊢ A ∈ V → [˙A / x]˙ x = B ↔ A = B