Metamath Proof Explorer


Theorem divgcdodd

Description: Either A / ( A gcd B ) is odd or B / ( A gcd B ) is odd. (Contributed by Scott Fenton, 19-Apr-2014)

Ref Expression
Assertion divgcdodd A B ¬ 2 A A gcd B ¬ 2 B A gcd B

Proof

Step Hyp Ref Expression
1 n2dvds1 ¬ 2 1
2 2z 2
3 nnz A A
4 nnz B B
5 gcddvds A B A gcd B A A gcd B B
6 3 4 5 syl2an A B A gcd B A A gcd B B
7 6 simpld A B A gcd B A
8 gcdnncl A B A gcd B
9 8 nnzd A B A gcd B
10 8 nnne0d A B A gcd B 0
11 3 adantr A B A
12 dvdsval2 A gcd B A gcd B 0 A A gcd B A A A gcd B
13 9 10 11 12 syl3anc A B A gcd B A A A gcd B
14 7 13 mpbid A B A A gcd B
15 6 simprd A B A gcd B B
16 4 adantl A B B
17 dvdsval2 A gcd B A gcd B 0 B A gcd B B B A gcd B
18 9 10 16 17 syl3anc A B A gcd B B B A gcd B
19 15 18 mpbid A B B A gcd B
20 dvdsgcdb 2 A A gcd B B A gcd B 2 A A gcd B 2 B A gcd B 2 A A gcd B gcd B A gcd B
21 2 14 19 20 mp3an2i A B 2 A A gcd B 2 B A gcd B 2 A A gcd B gcd B A gcd B
22 gcddiv A B A gcd B A gcd B A A gcd B B A gcd B A gcd B = A A gcd B gcd B A gcd B
23 11 16 8 6 22 syl31anc A B A gcd B A gcd B = A A gcd B gcd B A gcd B
24 8 nncnd A B A gcd B
25 24 10 dividd A B A gcd B A gcd B = 1
26 23 25 eqtr3d A B A A gcd B gcd B A gcd B = 1
27 26 breq2d A B 2 A A gcd B gcd B A gcd B 2 1
28 27 biimpd A B 2 A A gcd B gcd B A gcd B 2 1
29 21 28 sylbid A B 2 A A gcd B 2 B A gcd B 2 1
30 29 expdimp A B 2 A A gcd B 2 B A gcd B 2 1
31 1 30 mtoi A B 2 A A gcd B ¬ 2 B A gcd B
32 31 ex A B 2 A A gcd B ¬ 2 B A gcd B
33 imor 2 A A gcd B ¬ 2 B A gcd B ¬ 2 A A gcd B ¬ 2 B A gcd B
34 32 33 sylib A B ¬ 2 A A gcd B ¬ 2 B A gcd B