Metamath Proof Explorer


Theorem affineequiv

Description: Equivalence between two ways of expressing B as an affine combination of A and C . (Contributed by David Moews, 28-Feb-2017)

Ref Expression
Hypotheses affineequiv.a φA
affineequiv.b φB
affineequiv.c φC
affineequiv.d φD
Assertion affineequiv φB=DA+1DCCB=DCA

Proof

Step Hyp Ref Expression
1 affineequiv.a φA
2 affineequiv.b φB
3 affineequiv.c φC
4 affineequiv.d φD
5 4 3 mulcld φDC
6 4 1 mulcld φDA
7 3 5 6 subsubd φCDCDA=C-DC+DA
8 3 5 subcld φCDC
9 8 6 addcomd φC-DC+DA=DA+C-DC
10 7 9 eqtr2d φDA+C-DC=CDCDA
11 1cnd φ1
12 11 4 3 subdird φ1DC=1CDC
13 3 mullidd φ1C=C
14 13 oveq1d φ1CDC=CDC
15 12 14 eqtrd φ1DC=CDC
16 15 oveq2d φDA+1DC=DA+C-DC
17 3 2 subcld φCB
18 3 1 subcld φCA
19 4 18 mulcld φDCA
20 2 17 19 addsubassd φB+CB-DCA=B+CB-DCA
21 2 3 pncan3d φB+C-B=C
22 4 3 1 subdid φDCA=DCDA
23 21 22 oveq12d φB+CB-DCA=CDCDA
24 20 23 eqtr3d φB+CB-DCA=CDCDA
25 10 16 24 3eqtr4d φDA+1DC=B+CB-DCA
26 25 eqeq2d φB=DA+1DCB=B+CB-DCA
27 2 addridd φB+0=B
28 27 eqeq1d φB+0=B+CB-DCAB=B+CB-DCA
29 0cnd φ0
30 17 19 subcld φC-B-DCA
31 2 29 30 addcand φB+0=B+CB-DCA0=C-B-DCA
32 26 28 31 3bitr2d φB=DA+1DC0=C-B-DCA
33 eqcom 0=C-B-DCAC-B-DCA=0
34 32 33 bitrdi φB=DA+1DCC-B-DCA=0
35 17 19 subeq0ad φC-B-DCA=0CB=DCA
36 34 35 bitrd φB=DA+1DCCB=DCA