Theorem 2sb5rf 2503

Description: Reversed double substitution. Usage of this theorem is discouraged because it depends on ax-13 2403. (Contributed by NM, 3-Feb-2005.) (Revised by Mario Carneiro, 6-Oct-2016.) Remove distinct variable constraints. (Revised by Wolf Lammen, 28-Sep-2018.) (New usage is discouraged.)

Hypotheses

Ref	Expression
2sb5rf.1	⊢ Ⅎ𝑧𝜑
2sb5rf.2	⊢ Ⅎ𝑤𝜑

Assertion

Ref	Expression
2sb5rf	⊢ (𝜑 ↔ ∃𝑧∃𝑤((𝑧 = 𝑥 ∧ 𝑤 = 𝑦) ∧ [𝑧 / 𝑥][𝑤 / 𝑦]𝜑))

Distinct variable group:   𝑧,𝑤

Allowed substitution hints:   𝜑(𝑥, 𝑦, 𝑧, 𝑤)

Proof of Theorem 2sb5rf

Step	Hyp	Ref	Expression
1		2sb5rf.1	. . . 4 ⊢ Ⅎ𝑧𝜑
2	1	19.41 2270	. . 3 ⊢ (∃𝑧(∃𝑤(𝑧 = 𝑥 ∧ 𝑤 = 𝑦) ∧ 𝜑) ↔ (∃𝑧∃𝑤(𝑧 = 𝑥 ∧ 𝑤 = 𝑦) ∧ 𝜑))
3		2sb5rf.2	. . . . 5 ⊢ Ⅎ𝑤𝜑
4	3	19.41 2270	. . . 4 ⊢ (∃𝑤((𝑧 = 𝑥 ∧ 𝑤 = 𝑦) ∧ 𝜑) ↔ (∃𝑤(𝑧 = 𝑥 ∧ 𝑤 = 𝑦) ∧ 𝜑))
5	4	exbii 1877	. . 3 ⊢ (∃𝑧∃𝑤((𝑧 = 𝑥 ∧ 𝑤 = 𝑦) ∧ 𝜑) ↔ ∃𝑧(∃𝑤(𝑧 = 𝑥 ∧ 𝑤 = 𝑦) ∧ 𝜑))
6		2ax6e 2502	. . . 4 ⊢ ∃𝑧∃𝑤(𝑧 = 𝑥 ∧ 𝑤 = 𝑦)
7	6	biantrur 539	. . 3 ⊢ (𝜑 ↔ (∃𝑧∃𝑤(𝑧 = 𝑥 ∧ 𝑤 = 𝑦) ∧ 𝜑))
8	2, 5, 7	3bitr4ri 307	. 2 ⊢ (𝜑 ↔ ∃𝑧∃𝑤((𝑧 = 𝑥 ∧ 𝑤 = 𝑦) ∧ 𝜑))
9		sbequ12r 2287	. . . . 5 ⊢ (𝑧 = 𝑥 → ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ [𝑤 / 𝑦]𝜑))
10		sbequ12r 2287	. . . . 5 ⊢ (𝑤 = 𝑦 → ([𝑤 / 𝑦]𝜑 ↔ 𝜑))
11	9, 10	sylan9bb 518	. . . 4 ⊢ ((𝑧 = 𝑥 ∧ 𝑤 = 𝑦) → ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ 𝜑))
12	11	pm5.32i 584	. . 3 ⊢ (((𝑧 = 𝑥 ∧ 𝑤 = 𝑦) ∧ [𝑧 / 𝑥][𝑤 / 𝑦]𝜑) ↔ ((𝑧 = 𝑥 ∧ 𝑤 = 𝑦) ∧ 𝜑))
13	12	2exbii 1878	. 2 ⊢ (∃𝑧∃𝑤((𝑧 = 𝑥 ∧ 𝑤 = 𝑦) ∧ [𝑧 / 𝑥][𝑤 / 𝑦]𝜑) ↔ ∃𝑧∃𝑤((𝑧 = 𝑥 ∧ 𝑤 = 𝑦) ∧ 𝜑))
14	8, 13	bitr4i 281	1 ⊢ (𝜑 ↔ ∃𝑧∃𝑤((𝑧 = 𝑥 ∧ 𝑤 = 𝑦) ∧ [𝑧 / 𝑥][𝑤 / 𝑦]𝜑))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 400  ∃wex 1808  Ⅎwnf 1812  [wsb 2095

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-10 2175  ax-11 2191  ax-12 2212  ax-13 2403

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1572  df-ex 1809  df-nf 1813  df-sb 2096

This theorem is used by:  sbel2x  2505