Proof of Theorem conjmul
Step | Hyp | Ref
| Expression |
1 | | simpll 764 |
. . . . . . 7
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → 𝑃 ∈
ℂ) |
2 | | simprl 768 |
. . . . . . 7
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → 𝑄 ∈
ℂ) |
3 | | reccl 11640 |
. . . . . . . 8
⊢ ((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) → (1 / 𝑃) ∈
ℂ) |
4 | 3 | adantr 481 |
. . . . . . 7
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (1 / 𝑃) ∈
ℂ) |
5 | 1, 2, 4 | mul32d 11185 |
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · 𝑄) · (1 / 𝑃)) = ((𝑃 · (1 / 𝑃)) · 𝑄)) |
6 | | recid 11647 |
. . . . . . . 8
⊢ ((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) → (𝑃 · (1 / 𝑃)) = 1) |
7 | 6 | oveq1d 7290 |
. . . . . . 7
⊢ ((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) → ((𝑃 · (1 / 𝑃)) · 𝑄) = (1 · 𝑄)) |
8 | 7 | adantr 481 |
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · (1 / 𝑃)) · 𝑄) = (1 · 𝑄)) |
9 | | mulid2 10974 |
. . . . . . 7
⊢ (𝑄 ∈ ℂ → (1
· 𝑄) = 𝑄) |
10 | 9 | ad2antrl 725 |
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (1 ·
𝑄) = 𝑄) |
11 | 5, 8, 10 | 3eqtrd 2782 |
. . . . 5
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · 𝑄) · (1 / 𝑃)) = 𝑄) |
12 | | reccl 11640 |
. . . . . . . 8
⊢ ((𝑄 ∈ ℂ ∧ 𝑄 ≠ 0) → (1 / 𝑄) ∈
ℂ) |
13 | 12 | adantl 482 |
. . . . . . 7
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (1 / 𝑄) ∈
ℂ) |
14 | 1, 2, 13 | mulassd 10998 |
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · 𝑄) · (1 / 𝑄)) = (𝑃 · (𝑄 · (1 / 𝑄)))) |
15 | | recid 11647 |
. . . . . . . 8
⊢ ((𝑄 ∈ ℂ ∧ 𝑄 ≠ 0) → (𝑄 · (1 / 𝑄)) = 1) |
16 | 15 | oveq2d 7291 |
. . . . . . 7
⊢ ((𝑄 ∈ ℂ ∧ 𝑄 ≠ 0) → (𝑃 · (𝑄 · (1 / 𝑄))) = (𝑃 · 1)) |
17 | 16 | adantl 482 |
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (𝑃 · (𝑄 · (1 / 𝑄))) = (𝑃 · 1)) |
18 | | mulid1 10973 |
. . . . . . 7
⊢ (𝑃 ∈ ℂ → (𝑃 · 1) = 𝑃) |
19 | 18 | ad2antrr 723 |
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (𝑃 · 1) = 𝑃) |
20 | 14, 17, 19 | 3eqtrd 2782 |
. . . . 5
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · 𝑄) · (1 / 𝑄)) = 𝑃) |
21 | 11, 20 | oveq12d 7293 |
. . . 4
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (((𝑃 · 𝑄) · (1 / 𝑃)) + ((𝑃 · 𝑄) · (1 / 𝑄))) = (𝑄 + 𝑃)) |
22 | | mulcl 10955 |
. . . . . 6
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → (𝑃 · 𝑄) ∈ ℂ) |
23 | 22 | ad2ant2r 744 |
. . . . 5
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (𝑃 · 𝑄) ∈ ℂ) |
24 | 23, 4, 13 | adddid 10999 |
. . . 4
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = (((𝑃 · 𝑄) · (1 / 𝑃)) + ((𝑃 · 𝑄) · (1 / 𝑄)))) |
25 | | addcom 11161 |
. . . . 5
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → (𝑃 + 𝑄) = (𝑄 + 𝑃)) |
26 | 25 | ad2ant2r 744 |
. . . 4
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (𝑃 + 𝑄) = (𝑄 + 𝑃)) |
27 | 21, 24, 26 | 3eqtr4d 2788 |
. . 3
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = (𝑃 + 𝑄)) |
28 | 22 | mulid1d 10992 |
. . . 4
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 · 𝑄) · 1) = (𝑃 · 𝑄)) |
29 | 28 | ad2ant2r 744 |
. . 3
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · 𝑄) · 1) = (𝑃 · 𝑄)) |
30 | 27, 29 | eqeq12d 2754 |
. 2
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = ((𝑃 · 𝑄) · 1) ↔ (𝑃 + 𝑄) = (𝑃 · 𝑄))) |
31 | | addcl 10953 |
. . . 4
⊢ (((1 /
𝑃) ∈ ℂ ∧ (1
/ 𝑄) ∈ ℂ) →
((1 / 𝑃) + (1 / 𝑄)) ∈
ℂ) |
32 | 3, 12, 31 | syl2an 596 |
. . 3
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((1 / 𝑃) + (1 / 𝑄)) ∈ ℂ) |
33 | | mulne0 11617 |
. . 3
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (𝑃 · 𝑄) ≠ 0) |
34 | | ax-1cn 10929 |
. . . 4
⊢ 1 ∈
ℂ |
35 | | mulcan 11612 |
. . . 4
⊢ ((((1 /
𝑃) + (1 / 𝑄)) ∈ ℂ ∧ 1 ∈ ℂ
∧ ((𝑃 · 𝑄) ∈ ℂ ∧ (𝑃 · 𝑄) ≠ 0)) → (((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = ((𝑃 · 𝑄) · 1) ↔ ((1 / 𝑃) + (1 / 𝑄)) = 1)) |
36 | 34, 35 | mp3an2 1448 |
. . 3
⊢ ((((1 /
𝑃) + (1 / 𝑄)) ∈ ℂ ∧ ((𝑃 · 𝑄) ∈ ℂ ∧ (𝑃 · 𝑄) ≠ 0)) → (((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = ((𝑃 · 𝑄) · 1) ↔ ((1 / 𝑃) + (1 / 𝑄)) = 1)) |
37 | 32, 23, 33, 36 | syl12anc 834 |
. 2
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = ((𝑃 · 𝑄) · 1) ↔ ((1 / 𝑃) + (1 / 𝑄)) = 1)) |
38 | | eqcom 2745 |
. . . 4
⊢ ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ (𝑃 · 𝑄) = (𝑃 + 𝑄)) |
39 | | muleqadd 11619 |
. . . 4
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 · 𝑄) = (𝑃 + 𝑄) ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1)) |
40 | 38, 39 | bitrid 282 |
. . 3
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1)) |
41 | 40 | ad2ant2r 744 |
. 2
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1)) |
42 | 30, 37, 41 | 3bitr3d 309 |
1
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (((1 / 𝑃) + (1 / 𝑄)) = 1 ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1)) |