Metamath Proof Explorer


Theorem lcmineqlem17

Description: Inequality of 2^{2n}. (Contributed by metakunt, 29-Apr-2024)

Ref Expression
Hypothesis lcmineqlem17.1 φ N 0
Assertion lcmineqlem17 φ 2 2 N 2 N + 1 ( 2 N N)

Proof

Step Hyp Ref Expression
1 lcmineqlem17.1 φ N 0
2 2nn0 2 0
3 2 a1i φ 2 0
4 3 1 nn0mulcld φ 2 N 0
5 binom11 2 N 0 2 2 N = k = 0 2 N ( 2 N k)
6 4 5 syl φ 2 2 N = k = 0 2 N ( 2 N k)
7 fzfid φ 0 2 N Fin
8 4 adantr φ k 0 2 N 2 N 0
9 elfzelz k 0 2 N k
10 9 adantl φ k 0 2 N k
11 8 10 jca φ k 0 2 N 2 N 0 k
12 bccl 2 N 0 k ( 2 N k) 0
13 11 12 syl φ k 0 2 N ( 2 N k) 0
14 13 nn0red φ k 0 2 N ( 2 N k)
15 1 nn0zd φ N
16 bccl 2 N 0 N ( 2 N N) 0
17 4 15 16 syl2anc φ ( 2 N N) 0
18 17 nn0red φ ( 2 N N)
19 18 adantr φ k 0 2 N ( 2 N N)
20 bcmax N 0 k ( 2 N k) ( 2 N N)
21 1 9 20 syl2an φ k 0 2 N ( 2 N k) ( 2 N N)
22 7 14 19 21 fsumle φ k = 0 2 N ( 2 N k) k = 0 2 N ( 2 N N)
23 6 22 eqbrtrd φ 2 2 N k = 0 2 N ( 2 N N)
24 17 nn0cnd φ ( 2 N N)
25 fsumconst 0 2 N Fin ( 2 N N) k = 0 2 N ( 2 N N) = 0 2 N ( 2 N N)
26 7 24 25 syl2anc φ k = 0 2 N ( 2 N N) = 0 2 N ( 2 N N)
27 hashfz0 2 N 0 0 2 N = 2 N + 1
28 4 27 syl φ 0 2 N = 2 N + 1
29 28 oveq1d φ 0 2 N ( 2 N N) = 2 N + 1 ( 2 N N)
30 26 29 eqtrd φ k = 0 2 N ( 2 N N) = 2 N + 1 ( 2 N N)
31 23 30 breqtrd φ 2 2 N 2 N + 1 ( 2 N N)