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Metamath Proof Explorer

Theorem sbccow

Description: A composition law for class substitution. Version of sbcco with a disjoint variable condition, which requires fewer axioms. (Contributed by NM, 26-Sep-2003) Avoid ax-13 . (Revised by GG, 10-Jan-2024)

Ref Expression
Assertion sbccow ⊢ [˙A / y]˙ [˙y / x]˙ φ ↔ [˙A / x]˙ φ

Proof

Step Hyp Ref Expression
1 sbcex ⊢ [˙A / y]˙ [˙y / x]˙ φ → A ∈ V
2 sbcex ⊢ [˙A / x]˙ φ → A ∈ V
3 dfsbcq ⊢ z = A → [˙z / y]˙ [˙y / x]˙ φ ↔ [˙A / y]˙ [˙y / x]˙ φ
4 dfsbcq ⊢ z = A → [˙z / x]˙ φ ↔ [˙A / x]˙ φ
5 sbsbc ⊢ y x φ ↔ [˙y / x]˙ φ
6 5 sbbii ⊢ z y y x φ ↔ z y [˙y / x]˙ φ
7 sbco2vv ⊢ z y y x φ ↔ z x φ
8 sbsbc ⊢ z y [˙y / x]˙ φ ↔ [˙z / y]˙ [˙y / x]˙ φ
9 6 7 8 3bitr3ri ⊢ [˙z / y]˙ [˙y / x]˙ φ ↔ z x φ
10 sbsbc ⊢ z x φ ↔ [˙z / x]˙ φ
11 9 10 bitri ⊢ [˙z / y]˙ [˙y / x]˙ φ ↔ [˙z / x]˙ φ
12 3 4 11 vtoclbg ⊢ A ∈ V → [˙A / y]˙ [˙y / x]˙ φ ↔ [˙A / x]˙ φ
13 1 2 12 pm5.21nii ⊢ [˙A / y]˙ [˙y / x]˙ φ ↔ [˙A / x]˙ φ