Theorem inteq 1968

Description: Equality law for intersection.

Assertion

Ref	Expression
inteq	|- (A = B -> |^|A = |^|B)

Proof of Theorem inteq

Step	Hyp	Ref	Expression
1		raleq 1324	. . 3 |- (A = B -> (A.y e. A x e. y <-> A.y e. B x e. y))
2	1	biabdv 1183	. 2 |- (A = B -> {x | A.y e. A x e. y} = {x | A.y e. B x e. y})
3		dfint2 1967	. 2 |- |^|A = {x | A.y e. A x e. y}
4		dfint2 1967	. 2 |- |^|B = {x | A.y e. B x e. y}
5	2, 3, 4	3eqtr4g 1147	1 |- (A = B -> |^|A = |^|B)

Syntax hints:   -> wi 2   e. wel 803  {cab 1090   = wceq 1091  A.wral 1201  |^|cint 1965

This theorem is referenced by:  inteqi 1969  inteqd 1970  intex 1986  elreldm 2554  elxp5 2641  fundmen 3333  xpsnen 3339  mapunen 3397  fiint 3445  xpnnen 4927  shintclt 5295  chintclt 5297

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-int 1966

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