Theorem erscottd 35884

Description: Applying the Scott operation to equivalence classes preserves their equality property. (Contributed by BTernaryTau, 20-Oct-2026.)

Hypotheses

Ref	Expression
erscottd.1	⊢ (𝜑 → 𝑅 Er 𝑉)
erscottd.2	⊢ (𝜑 → 𝐴 ∈ 𝑉)

Assertion

Ref	Expression
erscottd	⊢ (𝜑 → (𝐴𝑅𝐵 ↔ Scott {𝑥 ∣ 𝑥𝑅𝐴} = Scott {𝑥 ∣ 𝑥𝑅𝐵}))

Distinct variable groups:   𝜑,𝑥   𝑥,𝐴   𝑥,𝐵   𝑥,𝑅

Allowed substitution hint:   𝑉(𝑥)

Proof of Theorem erscottd

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		erscottd.1	. . . . . . . 8 ⊢ (𝜑 → 𝑅 Er 𝑉)
2	1	3ad2ant1 1151	. . . . . . 7 ⊢ ((𝜑 ∧ 𝐴𝑅𝐵 ∧ 𝑥𝑅𝐴) → 𝑅 Er 𝑉)
3		simp3 1156	. . . . . . 7 ⊢ ((𝜑 ∧ 𝐴𝑅𝐵 ∧ 𝑥𝑅𝐴) → 𝑥𝑅𝐴)
4		simp2 1155	. . . . . . 7 ⊢ ((𝜑 ∧ 𝐴𝑅𝐵 ∧ 𝑥𝑅𝐴) → 𝐴𝑅𝐵)
5	2, 3, 4	ertrd 8772	. . . . . 6 ⊢ ((𝜑 ∧ 𝐴𝑅𝐵 ∧ 𝑥𝑅𝐴) → 𝑥𝑅𝐵)
6	5	3expia 1139	. . . . 5 ⊢ ((𝜑 ∧ 𝐴𝑅𝐵) → (𝑥𝑅𝐴 → 𝑥𝑅𝐵))
7	1	3ad2ant1 1151	. . . . . . 7 ⊢ ((𝜑 ∧ 𝐴𝑅𝐵 ∧ 𝑥𝑅𝐵) → 𝑅 Er 𝑉)
8		simp3 1156	. . . . . . 7 ⊢ ((𝜑 ∧ 𝐴𝑅𝐵 ∧ 𝑥𝑅𝐵) → 𝑥𝑅𝐵)
9		simp2 1155	. . . . . . 7 ⊢ ((𝜑 ∧ 𝐴𝑅𝐵 ∧ 𝑥𝑅𝐵) → 𝐴𝑅𝐵)
10	7, 8, 9	ertr4d 8775	. . . . . 6 ⊢ ((𝜑 ∧ 𝐴𝑅𝐵 ∧ 𝑥𝑅𝐵) → 𝑥𝑅𝐴)
11	10	3expia 1139	. . . . 5 ⊢ ((𝜑 ∧ 𝐴𝑅𝐵) → (𝑥𝑅𝐵 → 𝑥𝑅𝐴))
12	6, 11	impbid 215	. . . 4 ⊢ ((𝜑 ∧ 𝐴𝑅𝐵) → (𝑥𝑅𝐴 ↔ 𝑥𝑅𝐵))
13	12	abbidv 2829	. . 3 ⊢ ((𝜑 ∧ 𝐴𝑅𝐵) → {𝑥 ∣ 𝑥𝑅𝐴} = {𝑥 ∣ 𝑥𝑅𝐵})
14	13	scotteqd 9967	. 2 ⊢ ((𝜑 ∧ 𝐴𝑅𝐵) → Scott {𝑥 ∣ 𝑥𝑅𝐴} = Scott {𝑥 ∣ 𝑥𝑅𝐵})
15		erscottd.2	. . . . . . 7 ⊢ (𝜑 → 𝐴 ∈ 𝑉)
16	1, 15	erref 8776	. . . . . . 7 ⊢ (𝜑 → 𝐴𝑅𝐴)
17		breq1 5111	. . . . . . 7 ⊢ (𝑥 = 𝐴 → (𝑥𝑅𝐴 ↔ 𝐴𝑅𝐴))
18	15, 16, 17	elabd 3638	. . . . . 6 ⊢ (𝜑 → 𝐴 ∈ {𝑥 ∣ 𝑥𝑅𝐴})
19		n0i 4289	. . . . . 6 ⊢ (𝐴 ∈ {𝑥 ∣ 𝑥𝑅𝐴} → ¬ {𝑥 ∣ 𝑥𝑅𝐴} = ∅)
20	18, 19	syl 18	. . . . 5 ⊢ (𝜑 → ¬ {𝑥 ∣ 𝑥𝑅𝐴} = ∅)
21		scott0b 9974	. . . . 5 ⊢ ({𝑥 ∣ 𝑥𝑅𝐴} = ∅ ↔ Scott {𝑥 ∣ 𝑥𝑅𝐴} = ∅)
22	20, 21	sylnib 331	. . . 4 ⊢ (𝜑 → ¬ Scott {𝑥 ∣ 𝑥𝑅𝐴} = ∅)
23		disjeq0 4412	. . . . . . 7 ⊢ ((Scott {𝑥 ∣ 𝑥𝑅𝐴} ∩ Scott {𝑥 ∣ 𝑥𝑅𝐵}) = ∅ → (Scott {𝑥 ∣ 𝑥𝑅𝐴} = Scott {𝑥 ∣ 𝑥𝑅𝐵} ↔ (Scott {𝑥 ∣ 𝑥𝑅𝐴} = ∅ ∧ Scott {𝑥 ∣ 𝑥𝑅𝐵} = ∅)))
24	23	biimpcd 252	. . . . . 6 ⊢ (Scott {𝑥 ∣ 𝑥𝑅𝐴} = Scott {𝑥 ∣ 𝑥𝑅𝐵} → ((Scott {𝑥 ∣ 𝑥𝑅𝐴} ∩ Scott {𝑥 ∣ 𝑥𝑅𝐵}) = ∅ → (Scott {𝑥 ∣ 𝑥𝑅𝐴} = ∅ ∧ Scott {𝑥 ∣ 𝑥𝑅𝐵} = ∅)))
25		simpl 488	. . . . . . 7 ⊢ ((Scott {𝑥 ∣ 𝑥𝑅𝐴} = ∅ ∧ Scott {𝑥 ∣ 𝑥𝑅𝐵} = ∅) → Scott {𝑥 ∣ 𝑥𝑅𝐴} = ∅)
26	25	con3i 155	. . . . . 6 ⊢ (¬ Scott {𝑥 ∣ 𝑥𝑅𝐴} = ∅ → ¬ (Scott {𝑥 ∣ 𝑥𝑅𝐴} = ∅ ∧ Scott {𝑥 ∣ 𝑥𝑅𝐵} = ∅))
27	24, 26	nsyli 158	. . . . 5 ⊢ (Scott {𝑥 ∣ 𝑥𝑅𝐴} = Scott {𝑥 ∣ 𝑥𝑅𝐵} → (¬ Scott {𝑥 ∣ 𝑥𝑅𝐴} = ∅ → ¬ (Scott {𝑥 ∣ 𝑥𝑅𝐴} ∩ Scott {𝑥 ∣ 𝑥𝑅𝐵}) = ∅))
28		df-ne 2960	. . . . . 6 ⊢ ((Scott {𝑥 ∣ 𝑥𝑅𝐴} ∩ Scott {𝑥 ∣ 𝑥𝑅𝐵}) ≠ ∅ ↔ ¬ (Scott {𝑥 ∣ 𝑥𝑅𝐴} ∩ Scott {𝑥 ∣ 𝑥𝑅𝐵}) = ∅)
29		ndisj 4321	. . . . . 6 ⊢ ((Scott {𝑥 ∣ 𝑥𝑅𝐴} ∩ Scott {𝑥 ∣ 𝑥𝑅𝐵}) ≠ ∅ ↔ ∃𝑦(𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐴} ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐵}))
30	28, 29	bitr3i 280	. . . . 5 ⊢ (¬ (Scott {𝑥 ∣ 𝑥𝑅𝐴} ∩ Scott {𝑥 ∣ 𝑥𝑅𝐵}) = ∅ ↔ ∃𝑦(𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐴} ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐵}))
31	27, 30	imbitrdi 254	. . . 4 ⊢ (Scott {𝑥 ∣ 𝑥𝑅𝐴} = Scott {𝑥 ∣ 𝑥𝑅𝐵} → (¬ Scott {𝑥 ∣ 𝑥𝑅𝐴} = ∅ → ∃𝑦(𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐴} ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐵})))
32	22, 31	mpan9 516	. . 3 ⊢ ((𝜑 ∧ Scott {𝑥 ∣ 𝑥𝑅𝐴} = Scott {𝑥 ∣ 𝑥𝑅𝐵}) → ∃𝑦(𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐴} ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐵}))
33	1	3ad2ant1 1151	. . . . . . 7 ⊢ ((𝜑 ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐴} ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐵}) → 𝑅 Er 𝑉)
34		breq1 5111	. . . . . . . . 9 ⊢ (𝑥 = 𝑦 → (𝑥𝑅𝐴 ↔ 𝑦𝑅𝐴))
35	34	elscottab 9979	. . . . . . . 8 ⊢ (𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐴} → 𝑦𝑅𝐴)
36	35	3ad2ant2 1152	. . . . . . 7 ⊢ ((𝜑 ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐴} ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐵}) → 𝑦𝑅𝐴)
37		breq1 5111	. . . . . . . . 9 ⊢ (𝑥 = 𝑦 → (𝑥𝑅𝐵 ↔ 𝑦𝑅𝐵))
38	37	elscottab 9979	. . . . . . . 8 ⊢ (𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐵} → 𝑦𝑅𝐵)
39	38	3ad2ant3 1153	. . . . . . 7 ⊢ ((𝜑 ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐴} ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐵}) → 𝑦𝑅𝐵)
40	33, 36, 39	ertr3d 8774	. . . . . 6 ⊢ ((𝜑 ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐴} ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐵}) → 𝐴𝑅𝐵)
41	40	3expib 1140	. . . . 5 ⊢ (𝜑 → ((𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐴} ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐵}) → 𝐴𝑅𝐵))
42	41	exlimdv 1968	. . . 4 ⊢ (𝜑 → (∃𝑦(𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐴} ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐵}) → 𝐴𝑅𝐵))
43	42	adantr 486	. . 3 ⊢ ((𝜑 ∧ Scott {𝑥 ∣ 𝑥𝑅𝐴} = Scott {𝑥 ∣ 𝑥𝑅𝐵}) → (∃𝑦(𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐴} ∧ 𝑦 ∈ Scott {𝑥 ∣ 𝑥𝑅𝐵}) → 𝐴𝑅𝐵))
44	32, 43	mpd 16	. 2 ⊢ ((𝜑 ∧ Scott {𝑥 ∣ 𝑥𝑅𝐴} = Scott {𝑥 ∣ 𝑥𝑅𝐵}) → 𝐴𝑅𝐵)
45	14, 44	impbida 813	1 ⊢ (𝜑 → (𝐴𝑅𝐵 ↔ Scott {𝑥 ∣ 𝑥𝑅𝐴} = Scott {𝑥 ∣ 𝑥𝑅𝐵}))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∧ w3a 1103   = wceq 1570  ∃wex 1812   ∈ wcel 2147  {cab 2741   ≠ wne 2959   ∩ cin 4092  ∅c0 4281   class class class wbr 5108   Er wer 8752  Scott cscott 9965

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735  ax-sep 5255  ax-nul 5266  ax-pow 5334  ax-pr 5398  ax-un 7783

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3or 1104  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2102  df-mo 2567  df-eu 2597  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ne 2960  df-ral 3081  df-rex 3091  df-reu 3370  df-rab 3417  df-v 3456  df-sbc 3743  df-csb 3852  df-ss 3901  df-pss 4017  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-pw 4562  df-sn 4589  df-pr 4591  df-op 4841  df-uni 4873  df-int 4913  df-iun 4958  df-iin 4959  df-br 5109  df-opab 5174  df-mpt 5193  df-tr 5219  df-id 5553  df-eprel 5558  df-po 5566  df-so 5567  df-fr 5611  df-we 5613  df-xp 5664  df-rel 5665  df-cnv 5666  df-co 5667  df-dm 5668  df-rn 5669  df-res 5670  df-ima 5671  df-pred 6330  df-ord 6391  df-on 6392  df-lim 6393  df-suc 6394  df-iota 6520  df-fun 6568  df-fn 6569  df-f 6570  df-f1 6571  df-fo 6572  df-f1o 6573  df-fv 6574  df-ov 7454  df-om 7912  df-2nd 8036  df-frecs 8337  df-wrecs 8368  df-recs 8417  df-rdg 8456  df-er 8755  df-r1 9805  df-rank 9806  df-scott 9966

This theorem is used by:  ecscottd  35885