Theorem caoprdistrr 4073

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 4065	. 2 |- (CG(AFB)) = ((CGA)F(CGB))
6		oprex 3989	. . 3 |- (AFB) e. V
7		caoprd.com	. . 3 |- (xGy) = (yGx)
8	1, 6, 7	caoprcom 4059	. 2 |- (CG(AFB)) = ((AFB)GC)
9	1, 2, 7	caoprcom 4059	. . 3 |- (CGA) = (AGC)
10	1, 3, 7	caoprcom 4059	. . 3 |- (CGB) = (BGC)
11	9, 10	opreq12i 3979	. 2 |- ((CGA)F(CGB)) = ((AGC)F(BGC))
12	5, 8, 11	3eqtr3 1506	1 |- ((AFB)GC) = ((AGC)F(BGC))

Syntax hints:   = wceq 958   e. wcel 960  Vcvv 1814  (class class class)co 3969

This theorem is referenced by:  caoprdilem 4074  addcmpblnq 5064  ltapq 5088  prlem934a 5149  mulcmpblnrlem 5194  recexsrlem 5224  mulgt0sr 5226

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-11 969  ax-12 970  ax-13 971  ax-14 972  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462  ax-sep 2708  ax-pow 2748  ax-pr 2785  ax-un 2872

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-3an 779  df-ex 983  df-sb 1174  df-eu 1384  df-mo 1385  df-clab 1467  df-cleq 1472  df-clel 1475  df-ne 1590  df-v 1815  df-dif 2052  df-un 2053  df-in 2054  df-ss 2056  df-nul 2284  df-pw 2406  df-sn 2416  df-pr 2417  df-op 2420  df-uni 2508  df-br 2625  df-opab 2672  df-xp 3190  df-cnv 3192  df-dm 3194  df-rn 3195  df-res 3196  df-ima 3197  df-fv 3204  df-opr 3971

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