Theorem supeq1 6399

Description: Equality theorem for supremum. (Contributed by NM, 22-May-1999.)

Assertion

Ref	Expression
supeq1	⊢ (𝐵 = 𝐶 → sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅))

Proof of Theorem supeq1

Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		raleq 2549	. . . . 5 ⊢ (𝐵 = 𝐶 → (∀𝑦 ∈ 𝐵 ¬ 𝑥𝑅𝑦 ↔ ∀𝑦 ∈ 𝐶 ¬ 𝑥𝑅𝑦))
2		rexeq 2550	. . . . . . 7 ⊢ (𝐵 = 𝐶 → (∃𝑧 ∈ 𝐵 𝑦𝑅𝑧 ↔ ∃𝑧 ∈ 𝐶 𝑦𝑅𝑧))
3	2	imbi2d 228	. . . . . 6 ⊢ (𝐵 = 𝐶 → ((𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐵 𝑦𝑅𝑧) ↔ (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐶 𝑦𝑅𝑧)))
4	3	ralbidv 2368	. . . . 5 ⊢ (𝐵 = 𝐶 → (∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐵 𝑦𝑅𝑧) ↔ ∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐶 𝑦𝑅𝑧)))
5	1, 4	anbi12d 456	. . . 4 ⊢ (𝐵 = 𝐶 → ((∀𝑦 ∈ 𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐵 𝑦𝑅𝑧)) ↔ (∀𝑦 ∈ 𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐶 𝑦𝑅𝑧))))
6	5	rabbidv 2593	. . 3 ⊢ (𝐵 = 𝐶 → {𝑥 ∈ 𝐴 ∣ (∀𝑦 ∈ 𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐵 𝑦𝑅𝑧))} = {𝑥 ∈ 𝐴 ∣ (∀𝑦 ∈ 𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐶 𝑦𝑅𝑧))})
7	6	unieqd 3612	. 2 ⊢ (𝐵 = 𝐶 → ∪ {𝑥 ∈ 𝐴 ∣ (∀𝑦 ∈ 𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐵 𝑦𝑅𝑧))} = ∪ {𝑥 ∈ 𝐴 ∣ (∀𝑦 ∈ 𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐶 𝑦𝑅𝑧))})
8		df-sup 6397	. 2 ⊢ sup(𝐵, 𝐴, 𝑅) = ∪ {𝑥 ∈ 𝐴 ∣ (∀𝑦 ∈ 𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐵 𝑦𝑅𝑧))}
9		df-sup 6397	. 2 ⊢ sup(𝐶, 𝐴, 𝑅) = ∪ {𝑥 ∈ 𝐴 ∣ (∀𝑦 ∈ 𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐶 𝑦𝑅𝑧))}
10	7, 8, 9	3eqtr4g 2138	1 ⊢ (𝐵 = 𝐶 → sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 102   = wceq 1284  ∀wral 2348  ∃wrex 2349  {crab 2352  ∪ cuni 3601   class class class wbr 3785  supcsup 6395

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ral 2353  df-rex 2354  df-rab 2357  df-uni 3602  df-sup 6397

This theorem is referenced by:  supeq1d  6400  supeq1i  6401  infeq1  6424