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Metamath Proof Explorer

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 ) )