Metamath Proof Explorer


Theorem sbcco3g

Description: Composition of two substitutions. Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker sbcco3gw when possible. (Contributed by NM, 27-Nov-2005) (Revised by Mario Carneiro, 11-Nov-2016) (New usage is discouraged.)

Ref Expression
Hypothesis sbcco3g.1 ⊢ ( 𝑥 = 𝐴 → 𝐵 = 𝐶 )
Assertion sbcco3g ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝜑 ↔ [ 𝐶 / 𝑦 ] 𝜑 ) )

Proof

Step Hyp Ref Expression
1 sbcco3g.1 ⊢ ( 𝑥 = 𝐴 → 𝐵 = 𝐶 )
2 sbcnestg ⊢ ( 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 ) )