Theorem caoprdistrr 4067

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 4059	. 2 |- (CG(AFB)) = ((CGA)F(CGB))
6		oprex 3983	. . 3 |- (AFB) e. V
7		caoprd.com	. . 3 |- (xGy) = (yGx)
8	1, 6, 7	caoprcom 4053	. 2 |- (CG(AFB)) = ((AFB)GC)
9	1, 2, 7	caoprcom 4053	. . 3 |- (CGA) = (AGC)
10	1, 3, 7	caoprcom 4053	. . 3 |- (CGB) = (BGC)
11	9, 10	opreq12i 3973	. 2 |- ((CGA)F(CGB)) = ((AGC)F(BGC))
12	5, 8, 11	3eqtr3 1503	1 |- ((AFB)GC) = ((AGC)F(BGC))

Syntax hints:   = wceq 956   e. wcel 958  Vcvv 1811  (class class class)co 3963

This theorem is referenced by:  caoprdilem 4068  addcmpblnq 5052  ltapq 5076  prlem934a 5137  mulcmpblnrlem 5182  recexsrlem 5212  mulgt0sr 5214

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-11 967  ax-12 968  ax-13 969  ax-14 970  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218  ax-ext 1459  ax-sep 2703  ax-pow 2742  ax-pr 2779  ax-un 2866

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-3an 777  df-ex 981  df-sb 1172  df-eu 1382  df-mo 1383  df-clab 1464  df-cleq 1469  df-clel 1472  df-ne 1587  df-v 1812  df-dif 2049  df-un 2050  df-in 2051  df-ss 2053  df-nul 2281  df-pw 2402  df-sn 2412  df-pr 2413  df-op 2416  df-uni 2504  df-br 2620  df-opab 2667  df-xp 3184  df-cnv 3186  df-dm 3188  df-rn 3189  df-res 3190  df-ima 3191  df-fv 3198  df-opr 3965

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