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) (Revised by Gino Giotto, 26-Jan-2024)

Ref Expression
Hypothesis sbcco3gw.1 ( 𝑥 = 𝐴𝐵 = 𝐶 )
Assertion sbcco3gw ( 𝐴𝑉 → ( [ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝜑[ 𝐶 / 𝑦 ] 𝜑 ) )

Proof

Step Hyp Ref Expression
1 sbcco3gw.1 ( 𝑥 = 𝐴𝐵 = 𝐶 )
2 sbcnestgw ( 𝐴𝑉 → ( [ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝜑[ 𝐴 / 𝑥 𝐵 / 𝑦 ] 𝜑 ) )
3 elex ( 𝐴𝑉𝐴 ∈ V )
4 nfcvd ( 𝐴 ∈ V → 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 ( 𝐴𝑉 → ( [ 𝐴 / 𝑥 𝐵 / 𝑦 ] 𝜑[ 𝐶 / 𝑦 ] 𝜑 ) )
8 2 7 bitrd ( 𝐴𝑉 → ( [ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝜑[ 𝐶 / 𝑦 ] 𝜑 ) )