Theorem difeq1d 2725

Description: Deduction adding difference to the right in a class equality.

Hypothesis

Ref	Expression
difeq1d.1	|- (ph -> A = B)

Assertion

Ref	Expression
difeq1d	|- (ph -> (A \ C) = (B \ C))

Proof of Theorem difeq1d

Step	Hyp	Ref	Expression
1		difeq1d.1	. 2 |- (ph -> A = B)
2		difeq1 2717	. 2 |- (A = B -> (A \ C) = (B \ C))
3	1, 2	syl 12	1 |- (ph -> (A \ C) = (B \ C))

Syntax hints:   -> wi 3   = wceq 1298   \ cdif 2590

This theorem is referenced by:  dffv2 4734  phplem4 5605  unfilem3 5643  alephsuc3 8854  cldval 8942  iscncl 9047  drngi 9493  subcld 10254  topbnd 15408  isufil 15564  acdcg 15750  acdc3g 15751  acdc5g 15752  hmeocld 15900  txcld 15914  ishgrag 16291  hgralem 16292  pltval 16781  isdivrngNEW 17160  invrfval 17170  watomfval 17392

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-8 1306  ax-9 1307  ax-10 1308  ax-11 1309  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-16 1580  ax-11o 1588  ax-ext 1865

This theorem depends on definitions:  df-bi 164  df-or 241  df-an 242  df-ex 1327  df-sb 1536  df-clab 1872  df-cleq 1877  df-clel 1880  df-rab 2112  df-dif 2597

Copyright terms: Public domain