Metamath Proof Explorer


Theorem 2lgsoddprmlem3d

Description: Lemma 4 for 2lgsoddprmlem3 . (Contributed by AV, 20-Jul-2021)

Ref Expression
Assertion 2lgsoddprmlem3d ( ( ( 7 ↑ 2 ) − 1 ) / 8 ) = ( 2 · 3 )

Proof

Step Hyp Ref Expression
1 6cn 6 ∈ ℂ
2 8cn 8 ∈ ℂ
3 0re 0 ∈ ℝ
4 8pos 0 < 8
5 3 4 gtneii 8 ≠ 0
6 1 2 5 divcan4i ( ( 6 · 8 ) / 8 ) = 6
7 1 2 mulcli ( 6 · 8 ) ∈ ℂ
8 ax-1cn 1 ∈ ℂ
9 4p3e7 ( 4 + 3 ) = 7
10 9 eqcomi 7 = ( 4 + 3 )
11 10 oveq1i ( 7 ↑ 2 ) = ( ( 4 + 3 ) ↑ 2 )
12 4cn 4 ∈ ℂ
13 3cn 3 ∈ ℂ
14 12 13 binom2i ( ( 4 + 3 ) ↑ 2 ) = ( ( ( 4 ↑ 2 ) + ( 2 · ( 4 · 3 ) ) ) + ( 3 ↑ 2 ) )
15 sq4e2t8 ( 4 ↑ 2 ) = ( 2 · 8 )
16 2t4e8 ( 2 · 4 ) = 8
17 16 oveq1i ( ( 2 · 4 ) · 3 ) = ( 8 · 3 )
18 2cn 2 ∈ ℂ
19 18 12 13 mulassi ( ( 2 · 4 ) · 3 ) = ( 2 · ( 4 · 3 ) )
20 2 13 mulcomi ( 8 · 3 ) = ( 3 · 8 )
21 17 19 20 3eqtr3i ( 2 · ( 4 · 3 ) ) = ( 3 · 8 )
22 15 21 oveq12i ( ( 4 ↑ 2 ) + ( 2 · ( 4 · 3 ) ) ) = ( ( 2 · 8 ) + ( 3 · 8 ) )
23 18 13 2 adddiri ( ( 2 + 3 ) · 8 ) = ( ( 2 · 8 ) + ( 3 · 8 ) )
24 3p2e5 ( 3 + 2 ) = 5
25 13 18 24 addcomli ( 2 + 3 ) = 5
26 25 oveq1i ( ( 2 + 3 ) · 8 ) = ( 5 · 8 )
27 22 23 26 3eqtr2i ( ( 4 ↑ 2 ) + ( 2 · ( 4 · 3 ) ) ) = ( 5 · 8 )
28 sq3 ( 3 ↑ 2 ) = 9
29 df-9 9 = ( 8 + 1 )
30 28 29 eqtri ( 3 ↑ 2 ) = ( 8 + 1 )
31 27 30 oveq12i ( ( ( 4 ↑ 2 ) + ( 2 · ( 4 · 3 ) ) ) + ( 3 ↑ 2 ) ) = ( ( 5 · 8 ) + ( 8 + 1 ) )
32 5cn 5 ∈ ℂ
33 32 2 mulcli ( 5 · 8 ) ∈ ℂ
34 33 2 8 addassi ( ( ( 5 · 8 ) + 8 ) + 1 ) = ( ( 5 · 8 ) + ( 8 + 1 ) )
35 df-6 6 = ( 5 + 1 )
36 35 oveq1i ( 6 · 8 ) = ( ( 5 + 1 ) · 8 )
37 32 a1i ( 8 ∈ ℂ → 5 ∈ ℂ )
38 id ( 8 ∈ ℂ → 8 ∈ ℂ )
39 37 38 adddirp1d ( 8 ∈ ℂ → ( ( 5 + 1 ) · 8 ) = ( ( 5 · 8 ) + 8 ) )
40 2 39 ax-mp ( ( 5 + 1 ) · 8 ) = ( ( 5 · 8 ) + 8 )
41 36 40 eqtri ( 6 · 8 ) = ( ( 5 · 8 ) + 8 )
42 41 eqcomi ( ( 5 · 8 ) + 8 ) = ( 6 · 8 )
43 42 oveq1i ( ( ( 5 · 8 ) + 8 ) + 1 ) = ( ( 6 · 8 ) + 1 )
44 31 34 43 3eqtr2i ( ( ( 4 ↑ 2 ) + ( 2 · ( 4 · 3 ) ) ) + ( 3 ↑ 2 ) ) = ( ( 6 · 8 ) + 1 )
45 14 44 eqtri ( ( 4 + 3 ) ↑ 2 ) = ( ( 6 · 8 ) + 1 )
46 11 45 eqtri ( 7 ↑ 2 ) = ( ( 6 · 8 ) + 1 )
47 7 8 46 mvrraddi ( ( 7 ↑ 2 ) − 1 ) = ( 6 · 8 )
48 47 oveq1i ( ( ( 7 ↑ 2 ) − 1 ) / 8 ) = ( ( 6 · 8 ) / 8 )
49 2t3e6 ( 2 · 3 ) = 6
50 6 48 49 3eqtr4i ( ( ( 7 ↑ 2 ) − 1 ) / 8 ) = ( 2 · 3 )