Metamath Proof Explorer


Theorem fltoprmlem1

Description: Lemma 1 for fltoprm : Every positive integer is either a power of 2 or has an odd prime factor. (Contributed by AV, 15-Sep-2026)

Ref Expression
Assertion fltoprmlem1
|- ( N e. NN -> ( E. p e. Prime ( 2 < p /\ p || N ) \/ E. n e. NN0 N = ( 2 ^ n ) ) )

Proof

Step Hyp Ref Expression
1 oddprmdvds
 |-  ( ( N e. NN /\ -. E. n e. NN0 N = ( 2 ^ n ) ) -> E. p e. ( Prime \ { 2 } ) p || N )
2 rexdifsn
 |-  ( E. p e. ( Prime \ { 2 } ) p || N <-> E. p e. Prime ( p =/= 2 /\ p || N ) )
3 simpr
 |-  ( ( N e. NN /\ p e. Prime ) -> p e. Prime )
4 3 anim1i
 |-  ( ( ( N e. NN /\ p e. Prime ) /\ p =/= 2 ) -> ( p e. Prime /\ p =/= 2 ) )
5 eldifsn
 |-  ( p e. ( Prime \ { 2 } ) <-> ( p e. Prime /\ p =/= 2 ) )
6 4 5 sylibr
 |-  ( ( ( N e. NN /\ p e. Prime ) /\ p =/= 2 ) -> p e. ( Prime \ { 2 } ) )
7 oddprmgt2
 |-  ( p e. ( Prime \ { 2 } ) -> 2 < p )
8 6 7 syl
 |-  ( ( ( N e. NN /\ p e. Prime ) /\ p =/= 2 ) -> 2 < p )
9 8 ex
 |-  ( ( N e. NN /\ p e. Prime ) -> ( p =/= 2 -> 2 < p ) )
10 9 anim1d
 |-  ( ( N e. NN /\ p e. Prime ) -> ( ( p =/= 2 /\ p || N ) -> ( 2 < p /\ p || N ) ) )
11 10 reximdva
 |-  ( N e. NN -> ( E. p e. Prime ( p =/= 2 /\ p || N ) -> E. p e. Prime ( 2 < p /\ p || N ) ) )
12 2 11 biimtrid
 |-  ( N e. NN -> ( E. p e. ( Prime \ { 2 } ) p || N -> E. p e. Prime ( 2 < p /\ p || N ) ) )
13 12 adantr
 |-  ( ( N e. NN /\ -. E. n e. NN0 N = ( 2 ^ n ) ) -> ( E. p e. ( Prime \ { 2 } ) p || N -> E. p e. Prime ( 2 < p /\ p || N ) ) )
14 1 13 mpd
 |-  ( ( N e. NN /\ -. E. n e. NN0 N = ( 2 ^ n ) ) -> E. p e. Prime ( 2 < p /\ p || N ) )
15 14 orcd
 |-  ( ( N e. NN /\ -. E. n e. NN0 N = ( 2 ^ n ) ) -> ( E. p e. Prime ( 2 < p /\ p || N ) \/ E. n e. NN0 N = ( 2 ^ n ) ) )
16 15 ex
 |-  ( N e. NN -> ( -. E. n e. NN0 N = ( 2 ^ n ) -> ( E. p e. Prime ( 2 < p /\ p || N ) \/ E. n e. NN0 N = ( 2 ^ n ) ) ) )
17 olc
 |-  ( E. n e. NN0 N = ( 2 ^ n ) -> ( E. p e. Prime ( 2 < p /\ p || N ) \/ E. n e. NN0 N = ( 2 ^ n ) ) )
18 16 17 pm2.61d2
 |-  ( N e. NN -> ( E. p e. Prime ( 2 < p /\ p || N ) \/ E. n e. NN0 N = ( 2 ^ n ) ) )