Theorem difeq1 1582

Description: Equality theorem for class difference.

Assertion

Ref	Expression
difeq1	|- (A = B -> (A \ C) = (B \ C))

Proof of Theorem difeq1

Step	Hyp	Ref	Expression
1		eleq2 1150	. . . 4 |- (A = B -> (x e. A <-> x e. B))
2	1	anbi1d 469	. . 3 |- (A = B -> ((x e. A /\ -. x e. C) <-> (x e. B /\ -. x e. C)))
3	2	biabdv 1183	. 2 |- (A = B -> {x | (x e. A /\ -. x e. C)} = {x | (x e. B /\ -. x e. C)})
4		df-dif 1489	. 2 |- (A \ C) = {x | (x e. A /\ -. x e. C)}
5		df-dif 1489	. 2 |- (B \ C) = {x | (x e. B /\ -. x e. C)}
6	3, 4, 5	3eqtr4g 1147	1 |- (A = B -> (A \ C) = (B \ C))

Syntax hints:  -. wn 1   -> wi 2   /\ wa 196  {cab 1090   = wceq 1091   e. wcel 1092   \ cdif 1484

This theorem is referenced by:  difeq1i 1584  difeq1d 1587  kmlem8 3587  kmlem10 3589  kmlem11 3590

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-dif 1489

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