Metamath Proof Explorer


Theorem leexp1a

Description: Weak base ordering relationship for exponentiation. (Contributed by NM, 18-Dec-2005)

Ref Expression
Assertion leexp1a A B N 0 0 A A B A N B N

Proof

Step Hyp Ref Expression
1 oveq2 j = 0 A j = A 0
2 oveq2 j = 0 B j = B 0
3 1 2 breq12d j = 0 A j B j A 0 B 0
4 3 imbi2d j = 0 A B 0 A A B A j B j A B 0 A A B A 0 B 0
5 oveq2 j = k A j = A k
6 oveq2 j = k B j = B k
7 5 6 breq12d j = k A j B j A k B k
8 7 imbi2d j = k A B 0 A A B A j B j A B 0 A A B A k B k
9 oveq2 j = k + 1 A j = A k + 1
10 oveq2 j = k + 1 B j = B k + 1
11 9 10 breq12d j = k + 1 A j B j A k + 1 B k + 1
12 11 imbi2d j = k + 1 A B 0 A A B A j B j A B 0 A A B A k + 1 B k + 1
13 oveq2 j = N A j = A N
14 oveq2 j = N B j = B N
15 13 14 breq12d j = N A j B j A N B N
16 15 imbi2d j = N A B 0 A A B A j B j A B 0 A A B A N B N
17 recn A A
18 recn B B
19 exp0 A A 0 = 1
20 19 adantr A B A 0 = 1
21 1le1 1 1
22 20 21 eqbrtrdi A B A 0 1
23 exp0 B B 0 = 1
24 23 adantl A B B 0 = 1
25 22 24 breqtrrd A B A 0 B 0
26 17 18 25 syl2an A B A 0 B 0
27 26 adantr A B 0 A A B A 0 B 0
28 reexpcl A k 0 A k
29 28 ad4ant14 A B 0 A A B k 0 A k
30 simplll A B 0 A A B k 0 A
31 simpr A B 0 A A B k 0 k 0
32 simplrl A B 0 A A B k 0 0 A
33 expge0 A k 0 0 A 0 A k
34 30 31 32 33 syl3anc A B 0 A A B k 0 0 A k
35 reexpcl B k 0 B k
36 35 ad4ant24 A B 0 A A B k 0 B k
37 29 34 36 jca31 A B 0 A A B k 0 A k 0 A k B k
38 simpl A B A
39 simpl 0 A A B 0 A
40 38 39 anim12i A B 0 A A B A 0 A
41 40 adantr A B 0 A A B k 0 A 0 A
42 simpllr A B 0 A A B k 0 B
43 37 41 42 jca32 A B 0 A A B k 0 A k 0 A k B k A 0 A B
44 43 adantr A B 0 A A B k 0 A k B k A k 0 A k B k A 0 A B
45 simplrr A B 0 A A B k 0 A B
46 45 anim1ci A B 0 A A B k 0 A k B k A k B k A B
47 lemul12a A k 0 A k B k A 0 A B A k B k A B A k A B k B
48 44 46 47 sylc A B 0 A A B k 0 A k B k A k A B k B
49 expp1 A k 0 A k + 1 = A k A
50 17 49 sylan A k 0 A k + 1 = A k A
51 50 ad5ant14 A B 0 A A B k 0 A k B k A k + 1 = A k A
52 expp1 B k 0 B k + 1 = B k B
53 18 52 sylan B k 0 B k + 1 = B k B
54 53 ad5ant24 A B 0 A A B k 0 A k B k B k + 1 = B k B
55 48 51 54 3brtr4d A B 0 A A B k 0 A k B k A k + 1 B k + 1
56 55 ex A B 0 A A B k 0 A k B k A k + 1 B k + 1
57 56 expcom k 0 A B 0 A A B A k B k A k + 1 B k + 1
58 57 a2d k 0 A B 0 A A B A k B k A B 0 A A B A k + 1 B k + 1
59 4 8 12 16 27 58 nn0ind N 0 A B 0 A A B A N B N
60 59 exp4c N 0 A B 0 A A B A N B N
61 60 com3l A B N 0 0 A A B A N B N
62 61 3imp1 A B N 0 0 A A B A N B N