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Metamath Proof Explorer

Theorem sbcnestg

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

		Ref	Expression
	Assertion	sbcnestg	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝜑 ↔ [ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 / 𝑦 ] 𝜑 ) )

Proof

Step	Hyp	Ref	Expression
1		nfv	⊢ Ⅎ 𝑥 𝜑
2	1	ax-gen	⊢ ∀ 𝑦 Ⅎ 𝑥 𝜑
3		sbcnestgf	⊢ ( ( 𝐴 ∈ 𝑉 ∧ ∀ 𝑦 Ⅎ 𝑥 𝜑 ) → ( [ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝜑 ↔ [ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 / 𝑦 ] 𝜑 ) )
4	2 3	mpan2	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝜑 ↔ [ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 / 𝑦 ] 𝜑 ) )