Metamath Proof Explorer


Theorem pcfaclem

Description: Lemma for pcfac . (Contributed by Mario Carneiro, 20-May-2014)

Ref Expression
Assertion pcfaclem N 0 M N P N P M = 0

Proof

Step Hyp Ref Expression
1 nn0ge0 N 0 0 N
2 1 3ad2ant1 N 0 M N P 0 N
3 nn0re N 0 N
4 3 3ad2ant1 N 0 M N P N
5 prmnn P P
6 5 3ad2ant3 N 0 M N P P
7 eluznn0 N 0 M N M 0
8 7 3adant3 N 0 M N P M 0
9 6 8 nnexpcld N 0 M N P P M
10 9 nnred N 0 M N P P M
11 9 nngt0d N 0 M N P 0 < P M
12 ge0div N P M 0 < P M 0 N 0 N P M
13 4 10 11 12 syl3anc N 0 M N P 0 N 0 N P M
14 2 13 mpbid N 0 M N P 0 N P M
15 8 nn0red N 0 M N P M
16 eluzle M N N M
17 16 3ad2ant2 N 0 M N P N M
18 prmuz2 P P 2
19 18 3ad2ant3 N 0 M N P P 2
20 bernneq3 P 2 M 0 M < P M
21 19 8 20 syl2anc N 0 M N P M < P M
22 4 15 10 17 21 lelttrd N 0 M N P N < P M
23 9 nncnd N 0 M N P P M
24 23 mulid1d N 0 M N P P M 1 = P M
25 22 24 breqtrrd N 0 M N P N < P M 1
26 1red N 0 M N P 1
27 ltdivmul N 1 P M 0 < P M N P M < 1 N < P M 1
28 4 26 10 11 27 syl112anc N 0 M N P N P M < 1 N < P M 1
29 25 28 mpbird N 0 M N P N P M < 1
30 0p1e1 0 + 1 = 1
31 29 30 breqtrrdi N 0 M N P N P M < 0 + 1
32 4 9 nndivred N 0 M N P N P M
33 0z 0
34 flbi N P M 0 N P M = 0 0 N P M N P M < 0 + 1
35 32 33 34 sylancl N 0 M N P N P M = 0 0 N P M N P M < 0 + 1
36 14 31 35 mpbir2and N 0 M N P N P M = 0