Theorem 2uasbanh 37798

Description: Distribute the unabbreviated form of proper substitution in and out of a conjunction. 2uasbanh 37798 is derived from 2uasbanhVD 38169. (Contributed by Alan Sare, 31-May-2014.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypothesis

Ref	Expression
2uasbanh.1	⊢ (𝜒 ↔ (∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑) ∧ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜓)))

Assertion

Ref	Expression
2uasbanh	⊢ (∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (𝜑 ∧ 𝜓)) ↔ (∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑) ∧ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜓)))

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

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

Proof of Theorem 2uasbanh

Step	Hyp	Ref	Expression
1		simpl 472	. . . . 5 ⊢ (((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (𝜑 ∧ 𝜓)) → (𝑥 = 𝑢 ∧ 𝑦 = 𝑣))
2		simprl 790	. . . . 5 ⊢ (((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (𝜑 ∧ 𝜓)) → 𝜑)
3	1, 2	jca 553	. . . 4 ⊢ (((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (𝜑 ∧ 𝜓)) → ((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑))
4	3	2eximi 1753	. . 3 ⊢ (∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (𝜑 ∧ 𝜓)) → ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑))
5		simprr 792	. . . . 5 ⊢ (((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (𝜑 ∧ 𝜓)) → 𝜓)
6	1, 5	jca 553	. . . 4 ⊢ (((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (𝜑 ∧ 𝜓)) → ((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜓))
7	6	2eximi 1753	. . 3 ⊢ (∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (𝜑 ∧ 𝜓)) → ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜓))
8	4, 7	jca 553	. 2 ⊢ (∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (𝜑 ∧ 𝜓)) → (∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑) ∧ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜓)))
9		2uasbanh.1	. . 3 ⊢ (𝜒 ↔ (∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑) ∧ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜓)))
10	9	simplbi 475	. . . . . 6 ⊢ (𝜒 → ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑))
11		simpl 472	. . . . . . . . . 10 ⊢ (((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑) → (𝑥 = 𝑢 ∧ 𝑦 = 𝑣))
12	11	2eximi 1753	. . . . . . . . 9 ⊢ (∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑) → ∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣))
13	10, 12	syl 17	. . . . . . . 8 ⊢ (𝜒 → ∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣))
14		ax6e2ndeq 37796	. . . . . . . 8 ⊢ ((¬ ∀𝑥 𝑥 = 𝑦 ∨ 𝑢 = 𝑣) ↔ ∃𝑥∃𝑦(𝑥 = 𝑢 ∧ 𝑦 = 𝑣))
15	13, 14	sylibr 223	. . . . . . 7 ⊢ (𝜒 → (¬ ∀𝑥 𝑥 = 𝑦 ∨ 𝑢 = 𝑣))
16		2sb5nd 37797	. . . . . . 7 ⊢ ((¬ ∀𝑥 𝑥 = 𝑦 ∨ 𝑢 = 𝑣) → ([𝑢 / 𝑥][𝑣 / 𝑦]𝜑 ↔ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑)))
17	15, 16	syl 17	. . . . . 6 ⊢ (𝜒 → ([𝑢 / 𝑥][𝑣 / 𝑦]𝜑 ↔ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑)))
18	10, 17	mpbird 246	. . . . 5 ⊢ (𝜒 → [𝑢 / 𝑥][𝑣 / 𝑦]𝜑)
19	9	simprbi 479	. . . . . 6 ⊢ (𝜒 → ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜓))
20		2sb5nd 37797	. . . . . . 7 ⊢ ((¬ ∀𝑥 𝑥 = 𝑦 ∨ 𝑢 = 𝑣) → ([𝑢 / 𝑥][𝑣 / 𝑦]𝜓 ↔ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜓)))
21	15, 20	syl 17	. . . . . 6 ⊢ (𝜒 → ([𝑢 / 𝑥][𝑣 / 𝑦]𝜓 ↔ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜓)))
22	19, 21	mpbird 246	. . . . 5 ⊢ (𝜒 → [𝑢 / 𝑥][𝑣 / 𝑦]𝜓)
23		sban 2387	. . . . . . 7 ⊢ ([𝑣 / 𝑦](𝜑 ∧ 𝜓) ↔ ([𝑣 / 𝑦]𝜑 ∧ [𝑣 / 𝑦]𝜓))
24	23	sbbii 1874	. . . . . 6 ⊢ ([𝑢 / 𝑥][𝑣 / 𝑦](𝜑 ∧ 𝜓) ↔ [𝑢 / 𝑥]([𝑣 / 𝑦]𝜑 ∧ [𝑣 / 𝑦]𝜓))
25		sban 2387	. . . . . 6 ⊢ ([𝑢 / 𝑥]([𝑣 / 𝑦]𝜑 ∧ [𝑣 / 𝑦]𝜓) ↔ ([𝑢 / 𝑥][𝑣 / 𝑦]𝜑 ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜓))
26	24, 25	bitri 263	. . . . 5 ⊢ ([𝑢 / 𝑥][𝑣 / 𝑦](𝜑 ∧ 𝜓) ↔ ([𝑢 / 𝑥][𝑣 / 𝑦]𝜑 ∧ [𝑢 / 𝑥][𝑣 / 𝑦]𝜓))
27	18, 22, 26	sylanbrc 695	. . . 4 ⊢ (𝜒 → [𝑢 / 𝑥][𝑣 / 𝑦](𝜑 ∧ 𝜓))
28		2sb5nd 37797	. . . . 5 ⊢ ((¬ ∀𝑥 𝑥 = 𝑦 ∨ 𝑢 = 𝑣) → ([𝑢 / 𝑥][𝑣 / 𝑦](𝜑 ∧ 𝜓) ↔ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (𝜑 ∧ 𝜓))))
29	15, 28	syl 17	. . . 4 ⊢ (𝜒 → ([𝑢 / 𝑥][𝑣 / 𝑦](𝜑 ∧ 𝜓) ↔ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (𝜑 ∧ 𝜓))))
30	27, 29	mpbid 221	. . 3 ⊢ (𝜒 → ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (𝜑 ∧ 𝜓)))
31	9, 30	sylbir 224	. 2 ⊢ ((∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑) ∧ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜓)) → ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (𝜑 ∧ 𝜓)))
32	8, 31	impbii 198	1 ⊢ (∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ (𝜑 ∧ 𝜓)) ↔ (∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜑) ∧ ∃𝑥∃𝑦((𝑥 = 𝑢 ∧ 𝑦 = 𝑣) ∧ 𝜓)))

Syntax hints:  ¬ wn 3   ↔ wb 195   ∨ wo 382   ∧ wa 383  ∀wal 1473  ∃wex 1695  [wsb 1867

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-ne 2782  df-v 3175

This theorem is referenced by:  2uasban  37799