Theorem sbel2x 2508

Description: Elimination of double substitution. Usage of this theorem is discouraged because it depends on ax-13 2406. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Wolf Lammen, 29-Sep-2018.) (New usage is discouraged.)

Assertion

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

Distinct variable group:   𝑥,𝑦,𝜑

Allowed substitution hints:   𝜑(𝑧, 𝑤)

Proof of Theorem sbel2x

Step	Hyp	Ref	Expression
1		nfv 1947	. . 3 ⊢ Ⅎ𝑦𝜑
2		nfv 1947	. . 3 ⊢ Ⅎ𝑥𝜑
3	1, 2	2sb5rf 2506	. 2 ⊢ (𝜑 ↔ ∃𝑦∃𝑥((𝑦 = 𝑤 ∧ 𝑥 = 𝑧) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑))
4		ancom 466	. . . 4 ⊢ ((𝑦 = 𝑤 ∧ 𝑥 = 𝑧) ↔ (𝑥 = 𝑧 ∧ 𝑦 = 𝑤))
5	4	anbi1i 636	. . 3 ⊢ (((𝑦 = 𝑤 ∧ 𝑥 = 𝑧) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑) ↔ ((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑))
6	5	2exbii 1882	. 2 ⊢ (∃𝑦∃𝑥((𝑦 = 𝑤 ∧ 𝑥 = 𝑧) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑) ↔ ∃𝑦∃𝑥((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑))
7		excom 2200	. 2 ⊢ (∃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 300	1 ⊢ (𝜑 ↔ ∃𝑥∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401  ∃wex 1812  [wsb 2099

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-10 2179  ax-11 2195  ax-12 2216  ax-13 2406

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100

This theorem is used by: (None)