Theorem supeq1 2155

Description: Equality theorem for supremum.

Assertion

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

Proof of Theorem supeq1

Step	Hyp	Ref	Expression
1		raleq 1324	. . . . 5 ⊢ (B = C → (∀y ∈ B ¬ xRy ↔ ∀y ∈ C ¬ xRy))
2		rexeq 1325	. . . . . . 7 ⊢ (B = C → (∃z ∈ B yRz ↔ ∃z ∈ C yRz))
3	2	imbi2d 464	. . . . . 6 ⊢ (B = C → ((yRx → ∃z ∈ B yRz) ↔ (yRx → ∃z ∈ C yRz)))
4	3	biraldv 1219	. . . . 5 ⊢ (B = C → (∀y ∈ A (yRx → ∃z ∈ B yRz) ↔ ∀y ∈ A (yRx → ∃z ∈ C yRz)))
5	1, 4	anbi12d 476	. . . 4 ⊢ (B = C → ((∀y ∈ B ¬ xRy ∧ ∀y ∈ A (yRx → ∃z ∈ B yRz)) ↔ (∀y ∈ C ¬ xRy ∧ ∀y ∈ A (yRx → ∃z ∈ C yRz))))
6	5	birabsdv 1344	. . 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 1929	. 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 2154	. 2 ⊢ sup(B, A, R) = ∪{x ∈ A∣(∀y ∈ B ¬ xRy ∧ ∀y ∈ A (yRx → ∃z ∈ B yRz))}
9		df-sup 2154	. 2 ⊢ sup(C, A, R) = ∪{x ∈ A∣(∀y ∈ C ¬ xRy ∧ ∀y ∈ A (yRx → ∃z ∈ C yRz))}
10	7, 8, 9	3eqtr4g 1147	1 ⊢ (B = C → sup(B, A, R) = sup(C, A, R))

Syntax hints:  ¬ wn 1   → wi 2   ∧ wa 196   = wceq 1091  ∀wral 1201  ∃wrex 1202  {crab 1204  ∪cuni 1919   class class class wbr 2054  supcsup 2060

This theorem is referenced by:  sqrval 4729  sqr0 4730

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ral 1205  df-rex 1206  df-rab 1208  df-uni 1920  df-sup 2154

metamath.org