Metamath Proof Explorer


Theorem dvcmul

Description: The product rule when one argument is a constant. (Contributed by Mario Carneiro, 9-Aug-2014) (Revised by Mario Carneiro, 10-Feb-2015)

Ref Expression
Hypotheses dvcmul.s φ S
dvcmul.f φ F : X
dvcmul.a φ A
dvcmul.x φ X S
dvcmul.c φ C dom F S
Assertion dvcmul φ S × A × f F S C = A F S C

Proof

Step Hyp Ref Expression
1 dvcmul.s φ S
2 dvcmul.f φ F : X
3 dvcmul.a φ A
4 dvcmul.x φ X S
5 dvcmul.c φ C dom F S
6 fconst6g A S × A : S
7 3 6 syl φ S × A : S
8 ssidd φ S S
9 recnprss S S
10 1 9 syl φ S
11 10 2 4 dvbss φ dom F S X
12 11 5 sseldd φ C X
13 4 12 sseldd φ C S
14 fconst6g A × A :
15 3 14 syl φ × A :
16 ssidd φ
17 dvconst A D × A = × 0
18 3 17 syl φ D × A = × 0
19 18 dmeqd φ dom × A = dom × 0
20 c0ex 0 V
21 20 fconst × 0 : 0
22 21 fdmi dom × 0 =
23 19 22 eqtrdi φ dom × A =
24 10 23 sseqtrrd φ S dom × A
25 dvres3 S × A : S dom × A S D × A S = × A S
26 1 15 16 24 25 syl22anc φ S D × A S = × A S
27 xpssres S × A S = S × A
28 10 27 syl φ × A S = S × A
29 28 oveq2d φ S D × A S = S D S × A
30 18 reseq1d φ × A S = × 0 S
31 xpssres S × 0 S = S × 0
32 10 31 syl φ × 0 S = S × 0
33 30 32 eqtrd φ × A S = S × 0
34 26 29 33 3eqtr3d φ S D S × A = S × 0
35 20 fconst2 S × A S : S 0 S D S × A = S × 0
36 34 35 sylibr φ S × A S : S 0
37 36 fdmd φ dom S × A S = S
38 13 37 eleqtrrd φ C dom S × A S
39 7 8 2 4 1 38 5 dvmul φ S × A × f F S C = S × A S C F C + F S C S × A C
40 34 fveq1d φ S × A S C = S × 0 C
41 20 fvconst2 C S S × 0 C = 0
42 13 41 syl φ S × 0 C = 0
43 40 42 eqtrd φ S × A S C = 0
44 43 oveq1d φ S × A S C F C = 0 F C
45 2 12 ffvelrnd φ F C
46 45 mul02d φ 0 F C = 0
47 44 46 eqtrd φ S × A S C F C = 0
48 fvconst2g A C S S × A C = A
49 3 13 48 syl2anc φ S × A C = A
50 49 oveq2d φ F S C S × A C = F S C A
51 dvfg S F S : dom F S
52 1 51 syl φ F S : dom F S
53 52 5 ffvelrnd φ F S C
54 53 3 mulcomd φ F S C A = A F S C
55 50 54 eqtrd φ F S C S × A C = A F S C
56 47 55 oveq12d φ S × A S C F C + F S C S × A C = 0 + A F S C
57 3 53 mulcld φ A F S C
58 57 addid2d φ 0 + A F S C = A F S C
59 39 56 58 3eqtrd φ S × A × f F S C = A F S C