Metamath Proof Explorer


Theorem gausslemma2dlem0c

Description: Auxiliary lemma 3 for gausslemma2d . (Contributed by AV, 13-Jul-2021)

Ref Expression
Hypotheses gausslemma2dlem0a.p φ P 2
gausslemma2dlem0b.h H = P 1 2
Assertion gausslemma2dlem0c φ H ! gcd P = 1

Proof

Step Hyp Ref Expression
1 gausslemma2dlem0a.p φ P 2
2 gausslemma2dlem0b.h H = P 1 2
3 eldifi P 2 P
4 1 3 syl φ P
5 1 2 gausslemma2dlem0b φ H
6 5 nnnn0d φ H 0
7 4 6 jca φ P H 0
8 prmnn P P
9 nnre P P
10 peano2rem P P 1
11 9 10 syl P P 1
12 2re 2
13 12 a1i P 2
14 13 9 remulcld P 2 P
15 9 ltm1d P P 1 < P
16 nnnn0 P P 0
17 16 nn0ge0d P 0 P
18 1le2 1 2
19 18 a1i P 1 2
20 9 13 17 19 lemulge12d P P 2 P
21 11 9 14 15 20 ltletrd P P 1 < 2 P
22 2pos 0 < 2
23 12 22 pm3.2i 2 0 < 2
24 23 a1i P 2 0 < 2
25 ltdivmul P 1 P 2 0 < 2 P 1 2 < P P 1 < 2 P
26 11 9 24 25 syl3anc P P 1 2 < P P 1 < 2 P
27 21 26 mpbird P P 1 2 < P
28 3 8 27 3syl P 2 P 1 2 < P
29 1 28 syl φ P 1 2 < P
30 2 29 eqbrtrid φ H < P
31 prmndvdsfaclt P H 0 H < P ¬ P H !
32 7 30 31 sylc φ ¬ P H !
33 6 faccld φ H !
34 33 nnzd φ H !
35 nnz P P
36 3 8 35 3syl P 2 P
37 1 36 syl φ P
38 34 37 gcdcomd φ H ! gcd P = P gcd H !
39 38 eqeq1d φ H ! gcd P = 1 P gcd H ! = 1
40 coprm P H ! ¬ P H ! P gcd H ! = 1
41 4 34 40 syl2anc φ ¬ P H ! P gcd H ! = 1
42 39 41 bitr4d φ H ! gcd P = 1 ¬ P H !
43 32 42 mpbird φ H ! gcd P = 1