Theorem 2sb6

Description: Equivalence for double substitution. (Contributed by NM, 3-Feb-2005)

	Ref	Expression
Assertion	2sb6	⊢ ( [ 𝑧 / 𝑥 ] [ 𝑤 / 𝑦 ] 𝜑 ↔ ∀ 𝑥 ∀ 𝑦 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) → 𝜑 ) )

Proof

Step	Hyp	Ref	Expression
1		sb6	⊢ ( [ 𝑧 / 𝑥 ] [ 𝑤 / 𝑦 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝑧 → [ 𝑤 / 𝑦 ] 𝜑 ) )
2		19.21v	⊢ ( ∀ 𝑦 ( 𝑥 = 𝑧 → ( 𝑦 = 𝑤 → 𝜑 ) ) ↔ ( 𝑥 = 𝑧 → ∀ 𝑦 ( 𝑦 = 𝑤 → 𝜑 ) ) )
3		impexp	⊢ ( ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) → 𝜑 ) ↔ ( 𝑥 = 𝑧 → ( 𝑦 = 𝑤 → 𝜑 ) ) )
4	3	albii	⊢ ( ∀ 𝑦 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) → 𝜑 ) ↔ ∀ 𝑦 ( 𝑥 = 𝑧 → ( 𝑦 = 𝑤 → 𝜑 ) ) )
5		sb6	⊢ ( [ 𝑤 / 𝑦 ] 𝜑 ↔ ∀ 𝑦 ( 𝑦 = 𝑤 → 𝜑 ) )
6	5	imbi2i	⊢ ( ( 𝑥 = 𝑧 → [ 𝑤 / 𝑦 ] 𝜑 ) ↔ ( 𝑥 = 𝑧 → ∀ 𝑦 ( 𝑦 = 𝑤 → 𝜑 ) ) )
7	2 4 6	3bitr4ri	⊢ ( ( 𝑥 = 𝑧 → [ 𝑤 / 𝑦 ] 𝜑 ) ↔ ∀ 𝑦 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) → 𝜑 ) )
8	7	albii	⊢ ( ∀ 𝑥 ( 𝑥 = 𝑧 → [ 𝑤 / 𝑦 ] 𝜑 ) ↔ ∀ 𝑥 ∀ 𝑦 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) → 𝜑 ) )
9	1 8	bitri	⊢ ( [ 𝑧 / 𝑥 ] [ 𝑤 / 𝑦 ] 𝜑 ↔ ∀ 𝑥 ∀ 𝑦 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) → 𝜑 ) )