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 -> (A.y e. B -. xRy <-> A.y e. C -. xRy))
2		rexeq 1325	. . . . . . 7 |- (B = C -> (E.z e. B yRz <-> E.z e. C yRz))
3	2	imbi2d 464	. . . . . 6 |- (B = C -> ((yRx -> E.z e. B yRz) <-> (yRx -> E.z e. C yRz)))
4	3	biraldv 1219	. . . . 5 |- (B = C -> (A.y e. A (yRx -> E.z e. B yRz) <-> A.y e. A (yRx -> E.z e. C yRz)))
5	1, 4	anbi12d 476	. . . 4 |- (B = C -> ((A.y e. B -. xRy /\ A.y e. A (yRx -> E.z e. B yRz)) <-> (A.y e. C -. xRy /\ A.y e. A (yRx -> E.z e. C yRz))))
6	5	birabsdv 1344	. . 3 |- (B = C -> {x e. A | (A.y e. B -. xRy /\ A.y e. A (yRx -> E.z e. B yRz))} = {x e. A | (A.y e. C -. xRy /\ A.y e. A (yRx -> E.z e. C yRz))})
7	6	unieqd 1929	. 2 |- (B = C -> U.{x e. A | (A.y e. B -. xRy /\ A.y e. A (yRx -> E.z e. B yRz))} = U.{x e. A | (A.y e. C -. xRy /\ A.y e. A (yRx -> E.z e. C yRz))})
8		df-sup 2154	. 2 |- sup(B, A, R) = U.{x e. A | (A.y e. B -. xRy /\ A.y e. A (yRx -> E.z e. B yRz))}
9		df-sup 2154	. 2 |- sup(C, A, R) = U.{x e. A | (A.y e. C -. xRy /\ A.y e. A (yRx -> E.z e. 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  A.wral 1201  E.wrex 1202  {crab 1204  U.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

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