Theorem supeq1 6951

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 2661	. . . . 5 ⊢ (𝐵 = 𝐶 → (∀𝑦 ∈ 𝐵 ¬ 𝑥𝑅𝑦 ↔ ∀𝑦 ∈ 𝐶 ¬ 𝑥𝑅𝑦))
2		rexeq 2662	. . . . . . 7 ⊢ (𝐵 = 𝐶 → (∃𝑧 ∈ 𝐵 𝑦𝑅𝑧 ↔ ∃𝑧 ∈ 𝐶 𝑦𝑅𝑧))
3	2	imbi2d 229	. . . . . 6 ⊢ (𝐵 = 𝐶 → ((𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐵 𝑦𝑅𝑧) ↔ (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐶 𝑦𝑅𝑧)))
4	3	ralbidv 2466	. . . . 5 ⊢ (𝐵 = 𝐶 → (∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐵 𝑦𝑅𝑧) ↔ ∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐶 𝑦𝑅𝑧)))
5	1, 4	anbi12d 465	. . . 4 ⊢ (𝐵 = 𝐶 → ((∀𝑦 ∈ 𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐵 𝑦𝑅𝑧)) ↔ (∀𝑦 ∈ 𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐶 𝑦𝑅𝑧))))
6	5	rabbidv 2715	. . 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 3800	. 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 6949	. 2 ⊢ sup(𝐵, 𝐴, 𝑅) = ∪ {𝑥 ∈ 𝐴 ∣ (∀𝑦 ∈ 𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐵 𝑦𝑅𝑧))}
9		df-sup 6949	. 2 ⊢ sup(𝐶, 𝐴, 𝑅) = ∪ {𝑥 ∈ 𝐴 ∣ (∀𝑦 ∈ 𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐴 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐶 𝑦𝑅𝑧))}
10	7, 8, 9	3eqtr4g 2224	1 ⊢ (𝐵 = 𝐶 → sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   = wceq 1343  ∀wral 2444  ∃wrex 2445  {crab 2448  ∪ cuni 3789   class class class wbr 3982  supcsup 6947

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-ral 2449  df-rex 2450  df-rab 2453  df-uni 3790  df-sup 6949

This theorem is referenced by:  supeq1d  6952  supeq1i  6953  infeq1  6976