Theorem sbthlem5 9089

Description: Lemma for sbth 9095. (Contributed by NM, 22-Mar-1998.)

Hypotheses

Ref	Expression
sbthlem.1	⊢ 𝐴 ∈ V
sbthlem.2	⊢ 𝐷 = {𝑥 ∣ (𝑥 ⊆ 𝐴 ∧ (𝑔 “ (𝐵 ∖ (𝑓 “ 𝑥))) ⊆ (𝐴 ∖ 𝑥))}
sbthlem.3	⊢ 𝐻 = ((𝑓 ↾ ∪ 𝐷) ∪ (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷)))

Assertion

Ref	Expression
sbthlem5	⊢ ((dom 𝑓 = 𝐴 ∧ ran 𝑔 ⊆ 𝐴) → dom 𝐻 = 𝐴)

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝑥,𝐷   𝑥,𝑓   𝑥,𝑔   𝑥,𝐻

Allowed substitution hints:   𝐴(𝑓, 𝑔)   𝐵(𝑓, 𝑔)   𝐷(𝑓, 𝑔)   𝐻(𝑓, 𝑔)

Proof of Theorem sbthlem5

Step	Hyp	Ref	Expression
1		sbthlem.3	. . . . 5 ⊢ 𝐻 = ((𝑓 ↾ ∪ 𝐷) ∪ (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷)))
2	1	dmeqi 5888	. . . 4 ⊢ dom 𝐻 = dom ((𝑓 ↾ ∪ 𝐷) ∪ (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷)))
3		dmun 5894	. . . . 5 ⊢ dom ((𝑓 ↾ ∪ 𝐷) ∪ (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷))) = (dom (𝑓 ↾ ∪ 𝐷) ∪ dom (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷)))
4		dmres 6005	. . . . . 6 ⊢ dom (𝑓 ↾ ∪ 𝐷) = (∪ 𝐷 ∩ dom 𝑓)
5		dmres 6005	. . . . . . 7 ⊢ dom (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷)) = ((𝐴 ∖ ∪ 𝐷) ∩ dom ◡𝑔)
6		df-rn 5666	. . . . . . . . 9 ⊢ ran 𝑔 = dom ◡𝑔
7	6	eqcomi 2769	. . . . . . . 8 ⊢ dom ◡𝑔 = ran 𝑔
8	7	ineq2i 4163	. . . . . . 7 ⊢ ((𝐴 ∖ ∪ 𝐷) ∩ dom ◡𝑔) = ((𝐴 ∖ ∪ 𝐷) ∩ ran 𝑔)
9	5, 8	eqtri 2783	. . . . . 6 ⊢ dom (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷)) = ((𝐴 ∖ ∪ 𝐷) ∩ ran 𝑔)
10	4, 9	uneq12i 4113	. . . . 5 ⊢ (dom (𝑓 ↾ ∪ 𝐷) ∪ dom (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷))) = ((∪ 𝐷 ∩ dom 𝑓) ∪ ((𝐴 ∖ ∪ 𝐷) ∩ ran 𝑔))
11	3, 10	eqtri 2783	. . . 4 ⊢ dom ((𝑓 ↾ ∪ 𝐷) ∪ (◡𝑔 ↾ (𝐴 ∖ ∪ 𝐷))) = ((∪ 𝐷 ∩ dom 𝑓) ∪ ((𝐴 ∖ ∪ 𝐷) ∩ ran 𝑔))
12	2, 11	eqtri 2783	. . 3 ⊢ dom 𝐻 = ((∪ 𝐷 ∩ dom 𝑓) ∪ ((𝐴 ∖ ∪ 𝐷) ∩ ran 𝑔))
13		sbthlem.1	. . . . . . . . 9 ⊢ 𝐴 ∈ V
14		sbthlem.2	. . . . . . . . 9 ⊢ 𝐷 = {𝑥 ∣ (𝑥 ⊆ 𝐴 ∧ (𝑔 “ (𝐵 ∖ (𝑓 “ 𝑥))) ⊆ (𝐴 ∖ 𝑥))}
15	13, 14	sbthlem1 9085	. . . . . . . 8 ⊢ ∪ 𝐷 ⊆ (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))))
16		difss 4083	. . . . . . . 8 ⊢ (𝐴 ∖ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷)))) ⊆ 𝐴
17	15, 16	sstri 3940	. . . . . . 7 ⊢ ∪ 𝐷 ⊆ 𝐴
18		sseq2 3957	. . . . . . 7 ⊢ (dom 𝑓 = 𝐴 → (∪ 𝐷 ⊆ dom 𝑓 ↔ ∪ 𝐷 ⊆ 𝐴))
19	17, 18	mpbiri 261	. . . . . 6 ⊢ (dom 𝑓 = 𝐴 → ∪ 𝐷 ⊆ dom 𝑓)
20		dfss 3918	. . . . . 6 ⊢ (∪ 𝐷 ⊆ dom 𝑓 ↔ ∪ 𝐷 = (∪ 𝐷 ∩ dom 𝑓))
21	19, 20	sylib 221	. . . . 5 ⊢ (dom 𝑓 = 𝐴 → ∪ 𝐷 = (∪ 𝐷 ∩ dom 𝑓))
22	21	uneq1d 4114	. . . 4 ⊢ (dom 𝑓 = 𝐴 → (∪ 𝐷 ∪ (𝐴 ∖ ∪ 𝐷)) = ((∪ 𝐷 ∩ dom 𝑓) ∪ (𝐴 ∖ ∪ 𝐷)))
23	13, 14	sbthlem3 9087	. . . . . . 7 ⊢ (ran 𝑔 ⊆ 𝐴 → (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))) = (𝐴 ∖ ∪ 𝐷))
24		imassrn 6067	. . . . . . 7 ⊢ (𝑔 “ (𝐵 ∖ (𝑓 “ ∪ 𝐷))) ⊆ ran 𝑔
25	23, 24	eqsstrrdi 3976	. . . . . 6 ⊢ (ran 𝑔 ⊆ 𝐴 → (𝐴 ∖ ∪ 𝐷) ⊆ ran 𝑔)
26		dfss 3918	. . . . . 6 ⊢ ((𝐴 ∖ ∪ 𝐷) ⊆ ran 𝑔 ↔ (𝐴 ∖ ∪ 𝐷) = ((𝐴 ∖ ∪ 𝐷) ∩ ran 𝑔))
27	25, 26	sylib 221	. . . . 5 ⊢ (ran 𝑔 ⊆ 𝐴 → (𝐴 ∖ ∪ 𝐷) = ((𝐴 ∖ ∪ 𝐷) ∩ ran 𝑔))
28	27	uneq2d 4115	. . . 4 ⊢ (ran 𝑔 ⊆ 𝐴 → ((∪ 𝐷 ∩ dom 𝑓) ∪ (𝐴 ∖ ∪ 𝐷)) = ((∪ 𝐷 ∩ dom 𝑓) ∪ ((𝐴 ∖ ∪ 𝐷) ∩ ran 𝑔)))
29	22, 28	sylan9eq 2815	. . 3 ⊢ ((dom 𝑓 = 𝐴 ∧ ran 𝑔 ⊆ 𝐴) → (∪ 𝐷 ∪ (𝐴 ∖ ∪ 𝐷)) = ((∪ 𝐷 ∩ dom 𝑓) ∪ ((𝐴 ∖ ∪ 𝐷) ∩ ran 𝑔)))
30	12, 29	eqtr4id 2814	. 2 ⊢ ((dom 𝑓 = 𝐴 ∧ ran 𝑔 ⊆ 𝐴) → dom 𝐻 = (∪ 𝐷 ∪ (𝐴 ∖ ∪ 𝐷)))
31		undif 4438	. . 3 ⊢ (∪ 𝐷 ⊆ 𝐴 ↔ (∪ 𝐷 ∪ (𝐴 ∖ ∪ 𝐷)) = 𝐴)
32	17, 31	mpbi 233	. 2 ⊢ (∪ 𝐷 ∪ (𝐴 ∖ ∪ 𝐷)) = 𝐴
33	30, 32	eqtrdi 2811	1 ⊢ ((dom 𝑓 = 𝐴 ∧ ran 𝑔 ⊆ 𝐴) → dom 𝐻 = 𝐴)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2145  {cab 2738  Vcvv 3450   ∖ cdif 3896   ∪ cun 3897   ∩ cin 3898   ⊆ wss 3899  ∪ cuni 4867  ^◡ccnv 5654  dom cdm 5655  ran crn 5656   ↾ cres 5657   “ cima 5658

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-12 2213  ax-ext 2732  ax-sep 5251  ax-pr 5398

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5104  df-opab 5168  df-xp 5661  df-cnv 5663  df-dm 5665  df-rn 5666  df-res 5667  df-ima 5668

This theorem is used by:  sbthlem9  9093