Theorem sbel2x

Description: Elimination of double substitution. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 5-Aug-1993) (Proof shortened by Wolf Lammen, 29-Sep-2018) (New usage is discouraged.)

	Ref	Expression
Assertion	sbel2x	⊢ ( 𝜑 ↔ ∃ 𝑥 ∃ 𝑦 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) ∧ [ 𝑦 / 𝑤 ] [ 𝑥 / 𝑧 ] 𝜑 ) )

Proof

Step	Hyp	Ref	Expression
1		nfv	⊢ Ⅎ 𝑦 𝜑
2		nfv	⊢ Ⅎ 𝑥 𝜑
3	1 2	2sb5rf	⊢ ( 𝜑 ↔ ∃ 𝑦 ∃ 𝑥 ( ( 𝑦 = 𝑤 ∧ 𝑥 = 𝑧 ) ∧ [ 𝑦 / 𝑤 ] [ 𝑥 / 𝑧 ] 𝜑 ) )
4		ancom	⊢ ( ( 𝑦 = 𝑤 ∧ 𝑥 = 𝑧 ) ↔ ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) )
5	4	anbi1i	⊢ ( ( ( 𝑦 = 𝑤 ∧ 𝑥 = 𝑧 ) ∧ [ 𝑦 / 𝑤 ] [ 𝑥 / 𝑧 ] 𝜑 ) ↔ ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) ∧ [ 𝑦 / 𝑤 ] [ 𝑥 / 𝑧 ] 𝜑 ) )
6	5	2exbii	⊢ ( ∃ 𝑦 ∃ 𝑥 ( ( 𝑦 = 𝑤 ∧ 𝑥 = 𝑧 ) ∧ [ 𝑦 / 𝑤 ] [ 𝑥 / 𝑧 ] 𝜑 ) ↔ ∃ 𝑦 ∃ 𝑥 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) ∧ [ 𝑦 / 𝑤 ] [ 𝑥 / 𝑧 ] 𝜑 ) )
7		excom	⊢ ( ∃ 𝑦 ∃ 𝑥 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) ∧ [ 𝑦 / 𝑤 ] [ 𝑥 / 𝑧 ] 𝜑 ) ↔ ∃ 𝑥 ∃ 𝑦 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) ∧ [ 𝑦 / 𝑤 ] [ 𝑥 / 𝑧 ] 𝜑 ) )
8	3 6 7	3bitri	⊢ ( 𝜑 ↔ ∃ 𝑥 ∃ 𝑦 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) ∧ [ 𝑦 / 𝑤 ] [ 𝑥 / 𝑧 ] 𝜑 ) )