Proof of Theorem conjmulap
| Step | Hyp | Ref
 | Expression | 
| 1 |   | simpll 527 | 
. . . . . . 7
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → 𝑃 ∈ ℂ) | 
| 2 |   | simprl 529 | 
. . . . . . 7
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → 𝑄 ∈ ℂ) | 
| 3 |   | recclap 8706 | 
. . . . . . . 8
⊢ ((𝑃 ∈ ℂ ∧ 𝑃 # 0) → (1 / 𝑃) ∈
ℂ) | 
| 4 | 3 | adantr 276 | 
. . . . . . 7
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (1 / 𝑃) ∈ ℂ) | 
| 5 | 1, 2, 4 | mul32d 8179 | 
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑃)) = ((𝑃 · (1 / 𝑃)) · 𝑄)) | 
| 6 |   | recidap 8713 | 
. . . . . . . 8
⊢ ((𝑃 ∈ ℂ ∧ 𝑃 # 0) → (𝑃 · (1 / 𝑃)) = 1) | 
| 7 | 6 | oveq1d 5937 | 
. . . . . . 7
⊢ ((𝑃 ∈ ℂ ∧ 𝑃 # 0) → ((𝑃 · (1 / 𝑃)) · 𝑄) = (1 · 𝑄)) | 
| 8 | 7 | adantr 276 | 
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · (1 / 𝑃)) · 𝑄) = (1 · 𝑄)) | 
| 9 |   | mullid 8024 | 
. . . . . . 7
⊢ (𝑄 ∈ ℂ → (1
· 𝑄) = 𝑄) | 
| 10 | 9 | ad2antrl 490 | 
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (1 · 𝑄) = 𝑄) | 
| 11 | 5, 8, 10 | 3eqtrd 2233 | 
. . . . 5
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑃)) = 𝑄) | 
| 12 |   | recclap 8706 | 
. . . . . . . 8
⊢ ((𝑄 ∈ ℂ ∧ 𝑄 # 0) → (1 / 𝑄) ∈
ℂ) | 
| 13 | 12 | adantl 277 | 
. . . . . . 7
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (1 / 𝑄) ∈ ℂ) | 
| 14 | 1, 2, 13 | mulassd 8050 | 
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑄)) = (𝑃 · (𝑄 · (1 / 𝑄)))) | 
| 15 |   | recidap 8713 | 
. . . . . . . 8
⊢ ((𝑄 ∈ ℂ ∧ 𝑄 # 0) → (𝑄 · (1 / 𝑄)) = 1) | 
| 16 | 15 | oveq2d 5938 | 
. . . . . . 7
⊢ ((𝑄 ∈ ℂ ∧ 𝑄 # 0) → (𝑃 · (𝑄 · (1 / 𝑄))) = (𝑃 · 1)) | 
| 17 | 16 | adantl 277 | 
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · (𝑄 · (1 / 𝑄))) = (𝑃 · 1)) | 
| 18 |   | mulrid 8023 | 
. . . . . . 7
⊢ (𝑃 ∈ ℂ → (𝑃 · 1) = 𝑃) | 
| 19 | 18 | ad2antrr 488 | 
. . . . . 6
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · 1) = 𝑃) | 
| 20 | 14, 17, 19 | 3eqtrd 2233 | 
. . . . 5
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑄)) = 𝑃) | 
| 21 | 11, 20 | oveq12d 5940 | 
. . . 4
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (((𝑃 · 𝑄) · (1 / 𝑃)) + ((𝑃 · 𝑄) · (1 / 𝑄))) = (𝑄 + 𝑃)) | 
| 22 |   | mulcl 8006 | 
. . . . . 6
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → (𝑃 · 𝑄) ∈ ℂ) | 
| 23 | 22 | ad2ant2r 509 | 
. . . . 5
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · 𝑄) ∈ ℂ) | 
| 24 | 23, 4, 13 | adddid 8051 | 
. . . 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 8163 | 
. . . . 5
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → (𝑃 + 𝑄) = (𝑄 + 𝑃)) | 
| 26 | 25 | ad2ant2r 509 | 
. . . 4
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 + 𝑄) = (𝑄 + 𝑃)) | 
| 27 | 21, 24, 26 | 3eqtr4d 2239 | 
. . 3
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = (𝑃 + 𝑄)) | 
| 28 | 22 | mulridd 8043 | 
. . . 4
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 · 𝑄) · 1) = (𝑃 · 𝑄)) | 
| 29 | 28 | ad2ant2r 509 | 
. . 3
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · 1) = (𝑃 · 𝑄)) | 
| 30 | 27, 29 | eqeq12d 2211 | 
. 2
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = ((𝑃 · 𝑄) · 1) ↔ (𝑃 + 𝑄) = (𝑃 · 𝑄))) | 
| 31 |   | addcl 8004 | 
. . . 4
⊢ (((1 /
𝑃) ∈ ℂ ∧ (1
/ 𝑄) ∈ ℂ) →
((1 / 𝑃) + (1 / 𝑄)) ∈
ℂ) | 
| 32 | 3, 12, 31 | syl2an 289 | 
. . 3
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((1 / 𝑃) + (1 / 𝑄)) ∈ ℂ) | 
| 33 |   | mulap0 8681 | 
. . 3
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · 𝑄) # 0) | 
| 34 |   | ax-1cn 7972 | 
. . . 4
⊢ 1 ∈
ℂ | 
| 35 |   | mulcanap 8692 | 
. . . 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 1336 | 
. . 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 1247 | 
. 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 2198 | 
. . . 4
⊢ ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ (𝑃 · 𝑄) = (𝑃 + 𝑄)) | 
| 39 |   | muleqadd 8695 | 
. . . 4
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 · 𝑄) = (𝑃 + 𝑄) ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1)) | 
| 40 | 38, 39 | bitrid 192 | 
. . 3
⊢ ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1)) | 
| 41 | 40 | ad2ant2r 509 | 
. 2
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1)) | 
| 42 | 30, 37, 41 | 3bitr3d 218 | 
1
⊢ (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (((1 / 𝑃) + (1 / 𝑄)) = 1 ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1)) |