Theorem difeq1 3182

Description: Equality theorem for class difference. (Contributed by NM, 10-Feb-1997.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

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

Proof of Theorem difeq1

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		rabeq 2673	. 2 |- ( A = B -> { x e. A | -. x e. C } = { x e. B | -. x e. C } )
2		dfdif2 3074	. 2 |- ( A \ C ) = { x e. A | -. x e. C }
3		dfdif2 3074	. 2 |- ( B \ C ) = { x e. B | -. x e. C }
4	1, 2, 3	3eqtr4g 2195	1 |- ( A = B -> ( A \ C ) = ( B \ C ) )

Syntax hints:   -. wn 3   -> wi 4   = wceq 1331   e. wcel 1480   {crab 2418   \ cdif 3063

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 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-rab 2423  df-dif 3068

This theorem is referenced by:  difeq12  3184  difeq1i  3185  difeq1d  3188  uneqdifeqim  3443  diffitest  6774