Metamath Proof Explorer


Theorem remulcand

Description: Commuted version of remulcan2d without ax-mulcom . (Contributed by SN, 21-Feb-2024)

Ref Expression
Hypotheses remulcand.1 φA
remulcand.2 φB
remulcand.3 φC
remulcand.4 φC0
Assertion remulcand φCA=CBA=B

Proof

Step Hyp Ref Expression
1 remulcand.1 φA
2 remulcand.2 φB
3 remulcand.3 φC
4 remulcand.4 φC0
5 ax-rrecex CC0xCx=1
6 3 4 5 syl2anc φxCx=1
7 3 adantr φxC
8 7 adantr φxCx=1C
9 simplr φxCx=1x
10 simpr φxCx=1Cx=1
11 8 9 10 remulinvcom φxCx=1xC=1
12 11 ex φxCx=1xC=1
13 oveq2 CA=CBxCA=xCB
14 13 3ad2ant3 φxxC=1CA=CBxCA=xCB
15 simp2 φxxC=1CA=CBxC=1
16 15 oveq1d φxxC=1CA=CBxCA=1A
17 simp1r φxxC=1CA=CBx
18 17 recnd φxxC=1CA=CBx
19 7 3ad2ant1 φxxC=1CA=CBC
20 19 recnd φxxC=1CA=CBC
21 simp1l φxxC=1CA=CBφ
22 21 1 syl φxxC=1CA=CBA
23 22 recnd φxxC=1CA=CBA
24 18 20 23 mulassd φxxC=1CA=CBxCA=xCA
25 remullid A1A=A
26 22 25 syl φxxC=1CA=CB1A=A
27 16 24 26 3eqtr3d φxxC=1CA=CBxCA=A
28 15 oveq1d φxxC=1CA=CBxCB=1B
29 21 2 syl φxxC=1CA=CBB
30 29 recnd φxxC=1CA=CBB
31 18 20 30 mulassd φxxC=1CA=CBxCB=xCB
32 remullid B1B=B
33 29 32 syl φxxC=1CA=CB1B=B
34 28 31 33 3eqtr3d φxxC=1CA=CBxCB=B
35 14 27 34 3eqtr3d φxxC=1CA=CBA=B
36 35 3exp φxxC=1CA=CBA=B
37 12 36 syld φxCx=1CA=CBA=B
38 37 impr φxCx=1CA=CBA=B
39 6 38 rexlimddv φCA=CBA=B
40 oveq2 A=BCA=CB
41 39 40 impbid1 φCA=CBA=B