Proof of Theorem minusmodnep2tmod
Step | Hyp | Ref
| Expression |
1 | | elpri 4671 |
. . . . . 6
⊢ (𝐵 ∈ {1, 2} → (𝐵 = 1 ∨ 𝐵 = 2)) |
2 | | 5ndvds3 16479 |
. . . . . . . 8
⊢ ¬ 5
∥ 3 |
3 | | oveq2 7459 |
. . . . . . . . . 10
⊢ (𝐵 = 1 → (3 · 𝐵) = (3 ·
1)) |
4 | | 3t1e3 12463 |
. . . . . . . . . 10
⊢ (3
· 1) = 3 |
5 | 3, 4 | eqtrdi 2796 |
. . . . . . . . 9
⊢ (𝐵 = 1 → (3 · 𝐵) = 3) |
6 | 5 | breq2d 5179 |
. . . . . . . 8
⊢ (𝐵 = 1 → (5 ∥ (3
· 𝐵) ↔ 5
∥ 3)) |
7 | 2, 6 | mtbiri 327 |
. . . . . . 7
⊢ (𝐵 = 1 → ¬ 5 ∥ (3
· 𝐵)) |
8 | | 5ndvds6 16480 |
. . . . . . . 8
⊢ ¬ 5
∥ 6 |
9 | | oveq2 7459 |
. . . . . . . . . 10
⊢ (𝐵 = 2 → (3 · 𝐵) = (3 ·
2)) |
10 | | 3t2e6 12464 |
. . . . . . . . . 10
⊢ (3
· 2) = 6 |
11 | 9, 10 | eqtrdi 2796 |
. . . . . . . . 9
⊢ (𝐵 = 2 → (3 · 𝐵) = 6) |
12 | 11 | breq2d 5179 |
. . . . . . . 8
⊢ (𝐵 = 2 → (5 ∥ (3
· 𝐵) ↔ 5
∥ 6)) |
13 | 8, 12 | mtbiri 327 |
. . . . . . 7
⊢ (𝐵 = 2 → ¬ 5 ∥ (3
· 𝐵)) |
14 | 7, 13 | jaoi 856 |
. . . . . 6
⊢ ((𝐵 = 1 ∨ 𝐵 = 2) → ¬ 5 ∥ (3 ·
𝐵)) |
15 | 1, 14 | syl 17 |
. . . . 5
⊢ (𝐵 ∈ {1, 2} → ¬ 5
∥ (3 · 𝐵)) |
16 | | fzo13pr 13816 |
. . . . 5
⊢ (1..^3) =
{1, 2} |
17 | 15, 16 | eleq2s 2862 |
. . . 4
⊢ (𝐵 ∈ (1..^3) → ¬ 5
∥ (3 · 𝐵)) |
18 | 17 | adantl 481 |
. . 3
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → ¬ 5
∥ (3 · 𝐵)) |
19 | | 5nn 12384 |
. . . . . . 7
⊢ 5 ∈
ℕ |
20 | 19 | a1i 11 |
. . . . . 6
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → 5 ∈
ℕ) |
21 | | simpl 482 |
. . . . . 6
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → 𝐴 ∈
ℤ) |
22 | | 2z 12681 |
. . . . . . . 8
⊢ 2 ∈
ℤ |
23 | 22 | a1i 11 |
. . . . . . 7
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → 2 ∈
ℤ) |
24 | | elfzoelz 13727 |
. . . . . . . 8
⊢ (𝐵 ∈ (1..^3) → 𝐵 ∈
ℤ) |
25 | 24 | adantl 481 |
. . . . . . 7
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → 𝐵 ∈
ℤ) |
26 | 23, 25 | zmulcld 12760 |
. . . . . 6
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → (2
· 𝐵) ∈
ℤ) |
27 | | submodaddmod 47264 |
. . . . . 6
⊢ ((5
∈ ℕ ∧ (𝐴
∈ ℤ ∧ (2 · 𝐵) ∈ ℤ ∧ 𝐵 ∈ ℤ)) → (((𝐴 + (2 · 𝐵)) mod 5) = ((𝐴 − 𝐵) mod 5) ↔ ((𝐴 + ((2 · 𝐵) + 𝐵)) mod 5) = (𝐴 mod 5))) |
28 | 20, 21, 26, 25, 27 | syl13anc 1372 |
. . . . 5
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → (((𝐴 + (2 · 𝐵)) mod 5) = ((𝐴 − 𝐵) mod 5) ↔ ((𝐴 + ((2 · 𝐵) + 𝐵)) mod 5) = (𝐴 mod 5))) |
29 | | 2cnd 12376 |
. . . . . . . . . . . 12
⊢ (𝐵 ∈ (1..^3) → 2 ∈
ℂ) |
30 | 24 | zcnd 12755 |
. . . . . . . . . . . 12
⊢ (𝐵 ∈ (1..^3) → 𝐵 ∈
ℂ) |
31 | 29, 30 | adddirp1d 11319 |
. . . . . . . . . . 11
⊢ (𝐵 ∈ (1..^3) → ((2 + 1)
· 𝐵) = ((2 ·
𝐵) + 𝐵)) |
32 | 31 | eqcomd 2746 |
. . . . . . . . . 10
⊢ (𝐵 ∈ (1..^3) → ((2
· 𝐵) + 𝐵) = ((2 + 1) · 𝐵)) |
33 | 32 | adantl 481 |
. . . . . . . . 9
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → ((2
· 𝐵) + 𝐵) = ((2 + 1) · 𝐵)) |
34 | | 2p1e3 12440 |
. . . . . . . . . 10
⊢ (2 + 1) =
3 |
35 | 34 | oveq1i 7461 |
. . . . . . . . 9
⊢ ((2 + 1)
· 𝐵) = (3 ·
𝐵) |
36 | 33, 35 | eqtrdi 2796 |
. . . . . . . 8
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → ((2
· 𝐵) + 𝐵) = (3 · 𝐵)) |
37 | 36 | oveq2d 7467 |
. . . . . . 7
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → (𝐴 + ((2 · 𝐵) + 𝐵)) = (𝐴 + (3 · 𝐵))) |
38 | 37 | oveq1d 7466 |
. . . . . 6
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → ((𝐴 + ((2 · 𝐵) + 𝐵)) mod 5) = ((𝐴 + (3 · 𝐵)) mod 5)) |
39 | 38 | eqeq1d 2742 |
. . . . 5
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → (((𝐴 + ((2 · 𝐵) + 𝐵)) mod 5) = (𝐴 mod 5) ↔ ((𝐴 + (3 · 𝐵)) mod 5) = (𝐴 mod 5))) |
40 | 28, 39 | bitrd 279 |
. . . 4
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → (((𝐴 + (2 · 𝐵)) mod 5) = ((𝐴 − 𝐵) mod 5) ↔ ((𝐴 + (3 · 𝐵)) mod 5) = (𝐴 mod 5))) |
41 | | eqcom 2747 |
. . . . 5
⊢ (((𝐴 − 𝐵) mod 5) = ((𝐴 + (2 · 𝐵)) mod 5) ↔ ((𝐴 + (2 · 𝐵)) mod 5) = ((𝐴 − 𝐵) mod 5)) |
42 | 41 | a1i 11 |
. . . 4
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → (((𝐴 − 𝐵) mod 5) = ((𝐴 + (2 · 𝐵)) mod 5) ↔ ((𝐴 + (2 · 𝐵)) mod 5) = ((𝐴 − 𝐵) mod 5))) |
43 | | 3z 12682 |
. . . . . 6
⊢ 3 ∈
ℤ |
44 | 43 | a1i 11 |
. . . . 5
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → 3 ∈
ℤ) |
45 | | addmulmodb 16332 |
. . . . 5
⊢ ((5
∈ ℕ ∧ (𝐴
∈ ℤ ∧ 3 ∈ ℤ ∧ 𝐵 ∈ ℤ)) → (5 ∥ (3
· 𝐵) ↔ ((𝐴 + (3 · 𝐵)) mod 5) = (𝐴 mod 5))) |
46 | 20, 21, 44, 25, 45 | syl13anc 1372 |
. . . 4
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → (5
∥ (3 · 𝐵)
↔ ((𝐴 + (3 ·
𝐵)) mod 5) = (𝐴 mod 5))) |
47 | 40, 42, 46 | 3bitr4d 311 |
. . 3
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → (((𝐴 − 𝐵) mod 5) = ((𝐴 + (2 · 𝐵)) mod 5) ↔ 5 ∥ (3 · 𝐵))) |
48 | 18, 47 | mtbird 325 |
. 2
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → ¬
((𝐴 − 𝐵) mod 5) = ((𝐴 + (2 · 𝐵)) mod 5)) |
49 | 48 | neqned 2953 |
1
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ (1..^3)) → ((𝐴 − 𝐵) mod 5) ≠ ((𝐴 + (2 · 𝐵)) mod 5)) |