Proof of Theorem conjmul
| Step | Hyp | Ref
| Expression |
| 1 | | simpll 767 |
. . . . . . 7
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → 𝑃 ∈
ℂ) |
| 2 | | simprl 771 |
. . . . . . 7
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → 𝑄 ∈
ℂ) |
| 3 | | reccl 11929 |
. . . . . . . 8
⊢ ((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) → (1 / 𝑃) ∈
ℂ) |
| 4 | 3 | adantr 480 |
. . . . . . 7
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (1 / 𝑃) ∈
ℂ) |
| 5 | 1, 2, 4 | mul32d 11471 |
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · 𝑄) · (1 / 𝑃)) = ((𝑃 · (1 / 𝑃)) · 𝑄)) |
| 6 | | recid 11936 |
. . . . . . . 8
⊢ ((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) → (𝑃 · (1 / 𝑃)) = 1) |
| 7 | 6 | oveq1d 7446 |
. . . . . . 7
⊢ ((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) → ((𝑃 · (1 / 𝑃)) · 𝑄) = (1 · 𝑄)) |
| 8 | 7 | adantr 480 |
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · (1 / 𝑃)) · 𝑄) = (1 · 𝑄)) |
| 9 | | mullid 11260 |
. . . . . . 7
⊢ (𝑄 ∈ ℂ → (1
· 𝑄) = 𝑄) |
| 10 | 9 | ad2antrl 728 |
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (1 ·
𝑄) = 𝑄) |
| 11 | 5, 8, 10 | 3eqtrd 2781 |
. . . . 5
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · 𝑄) · (1 / 𝑃)) = 𝑄) |
| 12 | | reccl 11929 |
. . . . . . . 8
⊢ ((𝑄 ∈ ℂ ∧ 𝑄 ≠ 0) → (1 / 𝑄) ∈
ℂ) |
| 13 | 12 | adantl 481 |
. . . . . . 7
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (1 / 𝑄) ∈
ℂ) |
| 14 | 1, 2, 13 | mulassd 11284 |
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · 𝑄) · (1 / 𝑄)) = (𝑃 · (𝑄 · (1 / 𝑄)))) |
| 15 | | recid 11936 |
. . . . . . . 8
⊢ ((𝑄 ∈ ℂ ∧ 𝑄 ≠ 0) → (𝑄 · (1 / 𝑄)) = 1) |
| 16 | 15 | oveq2d 7447 |
. . . . . . 7
⊢ ((𝑄 ∈ ℂ ∧ 𝑄 ≠ 0) → (𝑃 · (𝑄 · (1 / 𝑄))) = (𝑃 · 1)) |
| 17 | 16 | adantl 481 |
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (𝑃 · (𝑄 · (1 / 𝑄))) = (𝑃 · 1)) |
| 18 | | mulrid 11259 |
. . . . . . 7
⊢ (𝑃 ∈ ℂ → (𝑃 · 1) = 𝑃) |
| 19 | 18 | ad2antrr 726 |
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (𝑃 · 1) = 𝑃) |
| 20 | 14, 17, 19 | 3eqtrd 2781 |
. . . . 5
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · 𝑄) · (1 / 𝑄)) = 𝑃) |
| 21 | 11, 20 | oveq12d 7449 |
. . . 4
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (((𝑃 · 𝑄) · (1 / 𝑃)) + ((𝑃 · 𝑄) · (1 / 𝑄))) = (𝑄 + 𝑃)) |
| 22 | | mulcl 11239 |
. . . . . 6
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → (𝑃 · 𝑄) ∈ ℂ) |
| 23 | 22 | ad2ant2r 747 |
. . . . 5
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (𝑃 · 𝑄) ∈ ℂ) |
| 24 | 23, 4, 13 | adddid 11285 |
. . . 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 11447 |
. . . . 5
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → (𝑃 + 𝑄) = (𝑄 + 𝑃)) |
| 26 | 25 | ad2ant2r 747 |
. . . 4
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (𝑃 + 𝑄) = (𝑄 + 𝑃)) |
| 27 | 21, 24, 26 | 3eqtr4d 2787 |
. . 3
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = (𝑃 + 𝑄)) |
| 28 | 22 | mulridd 11278 |
. . . 4
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 · 𝑄) · 1) = (𝑃 · 𝑄)) |
| 29 | 28 | ad2ant2r 747 |
. . 3
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → ((𝑃 · 𝑄) · 1) = (𝑃 · 𝑄)) |
| 30 | 27, 29 | eqeq12d 2753 |
. 2
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = ((𝑃 · 𝑄) · 1) ↔ (𝑃 + 𝑄) = (𝑃 · 𝑄))) |
| 31 | | addcl 11237 |
. . . 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 11905 |
. . 3
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 ≠ 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 ≠ 0)) → (𝑃 · 𝑄) ≠ 0) |
| 34 | | ax-1cn 11213 |
. . . 4
⊢ 1 ∈
ℂ |
| 35 | | mulcan 11900 |
. . . 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 1451 |
. . 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 837 |
. 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 2744 |
. . . 4
⊢ ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ (𝑃 · 𝑄) = (𝑃 + 𝑄)) |
| 39 | | muleqadd 11907 |
. . . 4
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 · 𝑄) = (𝑃 + 𝑄) ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1)) |
| 40 | 38, 39 | bitrid 283 |
. . 3
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1)) |
| 41 | 40 | ad2ant2r 747 |
. 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)) |