Metamath Proof Explorer


Theorem 41prothprmlem1

Description: Lemma 1 for 41prothprm . (Contributed by AV, 4-Jul-2020)

Ref Expression
Hypothesis 41prothprm.p 𝑃 = 4 1
Assertion 41prothprmlem1 ( ( 𝑃 − 1 ) / 2 ) = 2 0

Proof

Step Hyp Ref Expression
1 41prothprm.p 𝑃 = 4 1
2 dfdec10 4 1 = ( ( 1 0 · 4 ) + 1 )
3 1 2 eqtri 𝑃 = ( ( 1 0 · 4 ) + 1 )
4 3 oveq1i ( 𝑃 − 1 ) = ( ( ( 1 0 · 4 ) + 1 ) − 1 )
5 10nn 1 0 ∈ ℕ
6 5 nncni 1 0 ∈ ℂ
7 4cn 4 ∈ ℂ
8 6 7 mulcli ( 1 0 · 4 ) ∈ ℂ
9 pncan1 ( ( 1 0 · 4 ) ∈ ℂ → ( ( ( 1 0 · 4 ) + 1 ) − 1 ) = ( 1 0 · 4 ) )
10 8 9 ax-mp ( ( ( 1 0 · 4 ) + 1 ) − 1 ) = ( 1 0 · 4 )
11 4 10 eqtri ( 𝑃 − 1 ) = ( 1 0 · 4 )
12 11 oveq1i ( ( 𝑃 − 1 ) / 2 ) = ( ( 1 0 · 4 ) / 2 )
13 2cn 2 ∈ ℂ
14 2ne0 2 ≠ 0
15 6 7 13 14 divassi ( ( 1 0 · 4 ) / 2 ) = ( 1 0 · ( 4 / 2 ) )
16 4d2e2 ( 4 / 2 ) = 2
17 16 oveq2i ( 1 0 · ( 4 / 2 ) ) = ( 1 0 · 2 )
18 2nn0 2 ∈ ℕ0
19 18 dec0u ( 1 0 · 2 ) = 2 0
20 17 19 eqtri ( 1 0 · ( 4 / 2 ) ) = 2 0
21 15 20 eqtri ( ( 1 0 · 4 ) / 2 ) = 2 0
22 12 21 eqtri ( ( 𝑃 − 1 ) / 2 ) = 2 0