Metamath Proof Explorer


Theorem repr0

Description: There is exactly one representation with no elements (an empty sum), only for M = 0 . (Contributed by Thierry Arnoux, 2-Dec-2021)

Ref Expression
Hypotheses reprval.a φ A
reprval.m φ M
reprval.s φ S 0
Assertion repr0 φ A repr 0 M = if M = 0

Proof

Step Hyp Ref Expression
1 reprval.a φ A
2 reprval.m φ M
3 reprval.s φ S 0
4 0nn0 0 0
5 4 a1i φ 0 0
6 1 2 5 reprval φ A repr 0 M = c A 0 ..^ 0 | a 0 ..^ 0 c a = M
7 fzo0 0 ..^ 0 =
8 7 sumeq1i a 0 ..^ 0 c a = a c a
9 sum0 a c a = 0
10 8 9 eqtri a 0 ..^ 0 c a = 0
11 10 eqeq1i a 0 ..^ 0 c a = M 0 = M
12 11 a1i c = a 0 ..^ 0 c a = M 0 = M
13 0ex V
14 13 snid
15 nnex V
16 15 a1i φ V
17 16 1 ssexd φ A V
18 mapdm0 A V A =
19 17 18 syl φ A =
20 14 19 eleqtrrid φ A
21 7 oveq2i A 0 ..^ 0 = A
22 20 21 eleqtrrdi φ A 0 ..^ 0
23 22 adantr φ M = 0 A 0 ..^ 0
24 simpr φ M = 0 M = 0
25 24 eqcomd φ M = 0 0 = M
26 21 19 syl5eq φ A 0 ..^ 0 =
27 26 eleq2d φ c A 0 ..^ 0 c
28 27 biimpa φ c A 0 ..^ 0 c
29 elsni c c =
30 28 29 syl φ c A 0 ..^ 0 c =
31 30 ad4ant13 φ M = 0 c A 0 ..^ 0 a 0 ..^ 0 c a = M c =
32 12 23 25 31 rabeqsnd φ M = 0 c A 0 ..^ 0 | a 0 ..^ 0 c a = M =
33 32 eqcomd φ M = 0 = c A 0 ..^ 0 | a 0 ..^ 0 c a = M
34 10 a1i φ ¬ M = 0 c A 0 ..^ 0 a 0 ..^ 0 c a = 0
35 simplr φ ¬ M = 0 c A 0 ..^ 0 ¬ M = 0
36 35 neqned φ ¬ M = 0 c A 0 ..^ 0 M 0
37 36 necomd φ ¬ M = 0 c A 0 ..^ 0 0 M
38 34 37 eqnetrd φ ¬ M = 0 c A 0 ..^ 0 a 0 ..^ 0 c a M
39 38 neneqd φ ¬ M = 0 c A 0 ..^ 0 ¬ a 0 ..^ 0 c a = M
40 39 ralrimiva φ ¬ M = 0 c A 0 ..^ 0 ¬ a 0 ..^ 0 c a = M
41 rabeq0 c A 0 ..^ 0 | a 0 ..^ 0 c a = M = c A 0 ..^ 0 ¬ a 0 ..^ 0 c a = M
42 40 41 sylibr φ ¬ M = 0 c A 0 ..^ 0 | a 0 ..^ 0 c a = M =
43 42 eqcomd φ ¬ M = 0 = c A 0 ..^ 0 | a 0 ..^ 0 c a = M
44 33 43 ifeqda φ if M = 0 = c A 0 ..^ 0 | a 0 ..^ 0 c a = M
45 6 44 eqtr4d φ A repr 0 M = if M = 0