Metamath Proof Explorer


Theorem ex-exp

Description: Example for df-exp . (Contributed by AV, 4-Sep-2021)

Ref Expression
Assertion ex-exp 5 2 = 25 3 2 = 1 9

Proof

Step Hyp Ref Expression
1 df-5 5 = 4 + 1
2 1 oveq1i 5 2 = 4 + 1 2
3 4cn 4
4 binom21 4 4 + 1 2 = 4 2 + 2 4 + 1
5 3 4 ax-mp 4 + 1 2 = 4 2 + 2 4 + 1
6 2nn0 2 0
7 4nn0 4 0
8 4p1e5 4 + 1 = 5
9 sq4e2t8 4 2 = 2 8
10 8cn 8
11 2cn 2
12 8t2e16 8 2 = 16
13 10 11 12 mulcomli 2 8 = 16
14 9 13 eqtri 4 2 = 16
15 2t4e8 2 4 = 8
16 14 15 oveq12i 4 2 + 2 4 = 16 + 8
17 1nn0 1 0
18 6nn0 6 0
19 8nn0 8 0
20 eqid 16 = 16
21 1p1e2 1 + 1 = 2
22 6cn 6
23 8p6e14 8 + 6 = 14
24 10 22 23 addcomli 6 + 8 = 14
25 17 18 19 20 21 7 24 decaddci 16 + 8 = 24
26 16 25 eqtri 4 2 + 2 4 = 24
27 6 7 8 26 decsuc 4 2 + 2 4 + 1 = 25
28 5 27 eqtri 4 + 1 2 = 25
29 2 28 eqtri 5 2 = 25
30 3cn 3
31 30 negcli 3
32 expneg 3 2 0 3 2 = 1 3 2
33 31 6 32 mp2an 3 2 = 1 3 2
34 sqneg 3 3 2 = 3 2
35 30 34 ax-mp 3 2 = 3 2
36 sq3 3 2 = 9
37 35 36 eqtri 3 2 = 9
38 37 oveq2i 1 3 2 = 1 9
39 33 38 eqtri 3 2 = 1 9
40 29 39 pm3.2i 5 2 = 25 3 2 = 1 9