Metamath Proof Explorer


Theorem itg0

Description: The integral of anything on the empty set is zero. (Contributed by Mario Carneiro, 13-Aug-2014)

Ref Expression
Assertion itg0 A dx = 0

Proof

Step Hyp Ref Expression
1 eqid A i k = A i k
2 1 dfitg A dx = k = 0 3 i k 2 x if x 0 A i k A i k 0
3 ifan if x 0 A i k A i k 0 = if x if 0 A i k A i k 0 0
4 noel ¬ x
5 4 iffalsei if x if 0 A i k A i k 0 0 = 0
6 3 5 eqtri if x 0 A i k A i k 0 = 0
7 6 mpteq2i x if x 0 A i k A i k 0 = x 0
8 fconstmpt × 0 = x 0
9 7 8 eqtr4i x if x 0 A i k A i k 0 = × 0
10 9 fveq2i 2 x if x 0 A i k A i k 0 = 2 × 0
11 itg20 2 × 0 = 0
12 10 11 eqtri 2 x if x 0 A i k A i k 0 = 0
13 12 oveq2i i k 2 x if x 0 A i k A i k 0 = i k 0
14 ax-icn i
15 elfznn0 k 0 3 k 0
16 expcl i k 0 i k
17 14 15 16 sylancr k 0 3 i k
18 17 mul01d k 0 3 i k 0 = 0
19 13 18 syl5eq k 0 3 i k 2 x if x 0 A i k A i k 0 = 0
20 19 sumeq2i k = 0 3 i k 2 x if x 0 A i k A i k 0 = k = 0 3 0
21 fzfi 0 3 Fin
22 21 olci 0 3 0 0 3 Fin
23 sumz 0 3 0 0 3 Fin k = 0 3 0 = 0
24 22 23 ax-mp k = 0 3 0 = 0
25 20 24 eqtri k = 0 3 i k 2 x if x 0 A i k A i k 0 = 0
26 2 25 eqtri A dx = 0