Metamath Proof Explorer


Theorem absexp

Description: Absolute value of positive integer exponentiation. (Contributed by NM, 5-Jan-2006)

Ref Expression
Assertion absexp A N 0 A N = A N

Proof

Step Hyp Ref Expression
1 oveq2 j = 0 A j = A 0
2 1 fveq2d j = 0 A j = A 0
3 oveq2 j = 0 A j = A 0
4 2 3 eqeq12d j = 0 A j = A j A 0 = A 0
5 oveq2 j = k A j = A k
6 5 fveq2d j = k A j = A k
7 oveq2 j = k A j = A k
8 6 7 eqeq12d j = k A j = A j A k = A k
9 oveq2 j = k + 1 A j = A k + 1
10 9 fveq2d j = k + 1 A j = A k + 1
11 oveq2 j = k + 1 A j = A k + 1
12 10 11 eqeq12d j = k + 1 A j = A j A k + 1 = A k + 1
13 oveq2 j = N A j = A N
14 13 fveq2d j = N A j = A N
15 oveq2 j = N A j = A N
16 14 15 eqeq12d j = N A j = A j A N = A N
17 abs1 1 = 1
18 exp0 A A 0 = 1
19 18 fveq2d A A 0 = 1
20 abscl A A
21 20 recnd A A
22 21 exp0d A A 0 = 1
23 17 19 22 3eqtr4a A A 0 = A 0
24 oveq1 A k = A k A k A = A k A
25 24 adantl A k 0 A k = A k A k A = A k A
26 expp1 A k 0 A k + 1 = A k A
27 26 fveq2d A k 0 A k + 1 = A k A
28 expcl A k 0 A k
29 simpl A k 0 A
30 absmul A k A A k A = A k A
31 28 29 30 syl2anc A k 0 A k A = A k A
32 27 31 eqtrd A k 0 A k + 1 = A k A
33 32 adantr A k 0 A k = A k A k + 1 = A k A
34 expp1 A k 0 A k + 1 = A k A
35 21 34 sylan A k 0 A k + 1 = A k A
36 35 adantr A k 0 A k = A k A k + 1 = A k A
37 25 33 36 3eqtr4d A k 0 A k = A k A k + 1 = A k + 1
38 4 8 12 16 23 37 nn0indd A N 0 A N = A N