Theorem sbel2x 2465

Description: Elimination of double substitution. Usage of this theorem is discouraged because it depends on ax-13 2363. (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 1909	. . 3 ⊢ Ⅎ𝑦𝜑
2		nfv 1909	. . 3 ⊢ Ⅎ𝑥𝜑
3	1, 2	2sb5rf 2463	. 2 ⊢ (𝜑 ↔ ∃𝑦∃𝑥((𝑦 = 𝑤 ∧ 𝑥 = 𝑧) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑))
4		ancom 460	. . . 4 ⊢ ((𝑦 = 𝑤 ∧ 𝑥 = 𝑧) ↔ (𝑥 = 𝑧 ∧ 𝑦 = 𝑤))
5	4	anbi1i 623	. . 3 ⊢ (((𝑦 = 𝑤 ∧ 𝑥 = 𝑧) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑) ↔ ((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑))
6	5	2exbii 1843	. 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 2154	. 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 297	1 ⊢ (𝜑 ↔ ∃𝑥∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑))

Syntax hints:   ↔ wb 205   ∧ wa 395  ∃wex 1773  [wsb 2059

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-10 2129  ax-11 2146  ax-12 2163  ax-13 2363

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-tru 1536  df-ex 1774  df-nf 1778  df-sb 2060

This theorem is referenced by: (None)