Theorem caoprdistrr 4059

Description: Reverse distributive law.

Hypotheses

Ref	Expression
caoprd.1	|- A e. V
caoprd.2	|- B e. V
caoprd.3	|- C e. V
caoprd.com	|- (xGy) = (yGx)
caoprd.distr	|- (xG(yFz)) = ((xGy)F(xGz))

Assertion

Ref	Expression
caoprdistrr	|- ((AFB)GC) = ((AGC)F(BGC))

Distinct variable groups:   x,y,z,F   x,A,y,z   x,B,y,z   x,C,y,z   x,G,y,z

Proof of Theorem caoprdistrr

Step	Hyp	Ref	Expression
1		caoprd.3	. . 3 |- C e. V
2		caoprd.1	. . 3 |- A e. V
3		caoprd.2	. . 3 |- B e. V
4		caoprd.distr	. . 3 |- (xG(yFz)) = ((xGy)F(xGz))
5	1, 2, 3, 4	caoprdistr 4051	. 2 |- (CG(AFB)) = ((CGA)F(CGB))
6		oprex 3974	. . 3 |- (AFB) e. V
7		caoprd.com	. . 3 |- (xGy) = (yGx)
8	1, 6, 7	caoprcom 4045	. 2 |- (CG(AFB)) = ((AFB)GC)
9	1, 2, 7	caoprcom 4045	. . 3 |- (CGA) = (AGC)
10	1, 3, 7	caoprcom 4045	. . 3 |- (CGB) = (BGC)
11	9, 10	opreq12i 3964	. 2 |- ((CGA)F(CGB)) = ((AGC)F(BGC))
12	5, 8, 11	3eqtr3 1500	1 |- ((AFB)GC) = ((AGC)F(BGC))

Syntax hints:   = wceq 954   e. wcel 956  Vcvv 1807  (class class class)co 3954

This theorem is referenced by:  caoprdilem 4060  addcmpblnq 5032  ltapq 5056  prlem934a 5117  mulcmpblnrlem 5162  recexsrlem 5192  mulgt0sr 5194

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-8 962  ax-10 964  ax-11 965  ax-12 966  ax-13 967  ax-14 968  ax-17 969  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-16 1208  ax-11o 1216  ax-ext 1457  ax-sep 2698  ax-pow 2737  ax-pr 2774  ax-un 2861

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-3an 776  df-ex 979  df-sb 1170  df-eu 1380  df-mo 1381  df-clab 1462  df-cleq 1467  df-clel 1470  df-ne 1584  df-v 1808  df-dif 2045  df-un 2046  df-in 2047  df-ss 2049  df-nul 2277  df-pw 2398  df-sn 2408  df-pr 2409  df-op 2412  df-uni 2499  df-br 2615  df-opab 2662  df-xp 3179  df-cnv 3181  df-dm 3183  df-rn 3184  df-res 3185  df-ima 3186  df-fv 3193  df-opr 3956

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