Metamath Proof Explorer


Theorem frege55lem1c

Description: Necessary deduction regarding substitution of value in equality. (Contributed by RP, 24-Dec-2019)

Ref Expression
Assertion frege55lem1c ⊢ φ → [˙A / x]˙ x = B → φ → A = B

Proof

Step Hyp Ref Expression
1 df-sbc ⊢ [˙A / x]˙ x = B ↔ A ∈ x | x = B
2 eqeq1 ⊢ x = A → x = B ↔ A = B
3 2 elabg ⊢ A ∈ x | x = B → A ∈ x | x = B ↔ A = B
4 3 ibi ⊢ A ∈ x | x = B → A = B
5 1 4 sylbi ⊢ [˙A / x]˙ x = B → A = B
6 5 imim2i ⊢ φ → [˙A / x]˙ x = B → φ → A = B