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Metamath Proof Explorer

Theorem sbcco3gw

Description: Composition of two substitutions. Version of sbcco3g with a disjoint variable condition, which does not require ax-13 . (Contributed by NM, 27-Nov-2005) Avoid ax-13 . (Revised by GG, 26-Jan-2024)

Ref Expression
Hypothesis sbcco3gw.1 ⊢ x = A → B = C
Assertion sbcco3gw ⊢ A ∈ V → [˙A / x]˙ [˙B / y]˙ φ ↔ [˙C / y]˙ φ

Proof

Step Hyp Ref Expression
1 sbcco3gw.1 ⊢ x = A → B = C
2 sbcnestgw ⊢ A ∈ V → [˙A / x]˙ [˙B / y]˙ φ ↔ [˙⦋ A / x⦌ B / y]˙ φ
3 elex ⊢ A ∈ V → A ∈ V
4 nfcvd ⊢ A ∈ V → Ⅎ _ x C
5 4 1 csbiegf ⊢ A ∈ V → ⦋ A / x⦌ B = C
6 dfsbcq ⊢ ⦋ A / x⦌ B = C → [˙⦋ A / x⦌ B / y]˙ φ ↔ [˙C / y]˙ φ
7 3 5 6 3syl ⊢ A ∈ V → [˙⦋ A / x⦌ B / y]˙ φ ↔ [˙C / y]˙ φ
8 2 7 bitrd ⊢ A ∈ V → [˙A / x]˙ [˙B / y]˙ φ ↔ [˙C / y]˙ φ