Metamath Proof Explorer


Theorem lcmineqlem2

Description: Part of lcm inequality lemma, this part eventually shows that F times the least common multiple of 1 to n is an integer. (Contributed by metakunt, 29-Apr-2024)

Ref Expression
Hypotheses lcmineqlem2.1 F=01xM11xNMdx
lcmineqlem2.2 φN
lcmineqlem2.3 φM
lcmineqlem2.4 φMN
Assertion lcmineqlem2 φF=k=0NM1k(NMk)01xM1xkdx

Proof

Step Hyp Ref Expression
1 lcmineqlem2.1 F=01xM11xNMdx
2 lcmineqlem2.2 φN
3 lcmineqlem2.3 φM
4 lcmineqlem2.4 φMN
5 1 2 3 4 lcmineqlem1 φF=01xM1k=0NM1k(NMk)xkdx
6 eqid 01=01
7 fzfid φ0NMFin
8 0red φ0
9 1red φ1
10 unitsscn 01
11 resmpt 01xxM101=x01xM1
12 10 11 ax-mp xxM101=x01xM1
13 nnm1nn0 MM10
14 expcncf M10xxM1:cn
15 3 13 14 3syl φxxM1:cn
16 rescncf 01xxM1:cnxxM101:01cn
17 10 16 ax-mp xxM1:cnxxM101:01cn
18 15 17 syl φxxM101:01cn
19 12 18 eqeltrrid φx01xM1:01cn
20 elfznn0 k0NMk0
21 neg1cn 1
22 expcl 1k01k
23 21 22 mpan k01k
24 20 23 syl k0NM1k
25 24 adantl φk0NM1k
26 3 nnnn0d φM0
27 2 nnnn0d φN0
28 nn0sub M0N0MNNM0
29 26 27 28 syl2anc φMNNM0
30 4 29 mpbid φNM0
31 nn0z k0k
32 20 31 syl k0NMk
33 bccl NM0k(NMk)0
34 32 33 sylan2 NM0k0NM(NMk)0
35 30 34 sylan φk0NM(NMk)0
36 35 nn0cnd φk0NM(NMk)
37 25 36 mulcld φk0NM1k(NMk)
38 resmpt 01xxk01=x01xk
39 10 38 ax-mp xxk01=x01xk
40 expcncf k0xxk:cn
41 20 40 syl k0NMxxk:cn
42 rescncf 01xxk:cnxxk01:01cn
43 10 42 ax-mp xxk:cnxxk01:01cn
44 41 43 syl k0NMxxk01:01cn
45 39 44 eqeltrrid k0NMx01xk:01cn
46 45 adantl φk0NMx01xk:01cn
47 6 7 8 9 19 37 46 3factsumint φ01xM1k=0NM1k(NMk)xkdx=k=0NM1k(NMk)01xM1xkdx
48 5 47 eqtrd φF=k=0NM1k(NMk)01xM1xkdx