Theorem 2sb5nd 45542

Description: Equivalence for double substitution 2sb5 2312 without distinct 𝑥, 𝑦 requirement. 2sb5nd 45542 is derived from 2sb5ndVD 45891. (Contributed by Alan Sare, 30-Apr-2014.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
2sb5nd	⊢ ((¬ ∀𝑥 𝑥 = 𝑦 ∨ 𝑢 = 𝑣) → ([𝑢 / 𝑥][𝑣 / 𝑦]𝜑 ↔ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑)))

Distinct variable groups:   𝑥,𝑢   𝑦,𝑢   𝑥,𝑣   𝑦,𝑣

Allowed substitution hints:   𝜑(𝑥, 𝑦, 𝑣, 𝑢)

Proof of Theorem 2sb5nd

Step	Hyp	Ref	Expression
1		ax6e2ndeq 45541	. 2 ⊢ ((¬ ∀𝑥 𝑥 = 𝑦 ∨ 𝑢 = 𝑣) ↔ ∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣))
2		anabs5 676	. . . 4 ⊢ ((∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜑)) ↔ (∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜑))
3		2pm13.193 45534	. . . . . . . . 9 ⊢ (((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜑) ↔ ((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑))
4	3	exbii 1881	. . . . . . . 8 ⊢ (∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜑) ↔ ∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑))
5		nfs1v 2193	. . . . . . . . . 10 ⊢ Ⅎ𝑦[𝑣 / 𝑦]𝜑
6	5	nfsb 2553	. . . . . . . . 9 ⊢ Ⅎ𝑦[𝑢 / 𝑥][𝑣 / 𝑦]𝜑
7	6	19.41 2272	. . . . . . . 8 ⊢ (∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜑) ↔ (∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜑))
8	4, 7	bitr3i 280	. . . . . . 7 ⊢ (∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑) ↔ (∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜑))
9	8	exbii 1881	. . . . . 6 ⊢ (∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑) ↔ ∃𝑥(∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜑))
10		nfs1v 2193	. . . . . . 7 ⊢ Ⅎ𝑥[𝑢 / 𝑥][𝑣 / 𝑦]𝜑
11	10	19.41 2272	. . . . . 6 ⊢ (∃𝑥(∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜑) ↔ (∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜑))
12	9, 11	bitr2i 279	. . . . 5 ⊢ ((∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜑) ↔ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑))
13	12	anbi2i 635	. . . 4 ⊢ ((∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜑)) ↔ (∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑)))
14	2, 13	bitr3i 280	. . 3 ⊢ ((∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜑) ↔ (∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑)))
15		pm5.32 584	. . 3 ⊢ ((∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) → ([𝑢 / 𝑥][𝑣 / 𝑦]𝜑 ↔ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑))) ↔ ((∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜑) ↔ (∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑))))
16	14, 15	mpbir 234	. 2 ⊢ (∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣) → ([𝑢 / 𝑥][𝑣 / 𝑦]𝜑 ↔ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑)))
17	1, 16	sylbi 220	1 ⊢ ((¬ ∀𝑥 𝑥 = 𝑦 ∨ 𝑢 = 𝑣) → ([𝑢 / 𝑥][𝑣 / 𝑦]𝜑 ↔ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861  ∀wal 1568   = wceq 1570  ∃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-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-13 2402  ax-ext 2733

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  df-clab 2740  df-cleq 2753  df-clel 2836  df-ne 2957  df-v 3453

This theorem is used by:  2uasbanh  45543  2uasbanhVD  45892