Metamath Proof Explorer


Theorem 2lgsoddprmlem3b

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

Ref Expression
Assertion 2lgsoddprmlem3b ( ( ( 3 ↑ 2 ) − 1 ) / 8 ) = 1

Proof

Step Hyp Ref Expression
1 sq3 ( 3 ↑ 2 ) = 9
2 1 oveq1i ( ( 3 ↑ 2 ) − 1 ) = ( 9 − 1 )
3 9m1e8 ( 9 − 1 ) = 8
4 2 3 eqtri ( ( 3 ↑ 2 ) − 1 ) = 8
5 4 oveq1i ( ( ( 3 ↑ 2 ) − 1 ) / 8 ) = ( 8 / 8 )
6 8cn 8 ∈ ℂ
7 0re 0 ∈ ℝ
8 8pos 0 < 8
9 7 8 gtneii 8 ≠ 0
10 6 9 dividi ( 8 / 8 ) = 1
11 5 10 eqtri ( ( ( 3 ↑ 2 ) − 1 ) / 8 ) = 1