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Theorem conjmulap 9068
Description: Two numbers whose reciprocals sum to 1 are called "conjugates" and satisfy this relationship. (Contributed by Jim Kingdon, 26-Feb-2020.)
Assertion
Ref Expression
conjmulap (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (((1 / 𝑃) + (1 / 𝑄)) = 1 ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1))

Proof of Theorem conjmulap
StepHypRef Expression
1 simpll 531 . . . . . . 7 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → 𝑃 ∈ ℂ)
2 simprl 535 . . . . . . 7 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → 𝑄 ∈ ℂ)
3 recclap 9018 . . . . . . . 8 ((𝑃 ∈ ℂ ∧ 𝑃 # 0) → (1 / 𝑃) ∈ ℂ)
43adantr 276 . . . . . . 7 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (1 / 𝑃) ∈ ℂ)
51, 2, 4mul32d 8487 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑃)) = ((𝑃 · (1 / 𝑃)) · 𝑄))
6 recidap 9025 . . . . . . . 8 ((𝑃 ∈ ℂ ∧ 𝑃 # 0) → (𝑃 · (1 / 𝑃)) = 1)
76oveq1d 6103 . . . . . . 7 ((𝑃 ∈ ℂ ∧ 𝑃 # 0) → ((𝑃 · (1 / 𝑃)) · 𝑄) = (1 · 𝑄))
87adantr 276 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · (1 / 𝑃)) · 𝑄) = (1 · 𝑄))
9 mullid 8331 . . . . . . 7 (𝑄 ∈ ℂ → (1 · 𝑄) = 𝑄)
109ad2antrl 494 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (1 · 𝑄) = 𝑄)
115, 8, 103eqtrd 2275 . . . . 5 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑃)) = 𝑄)
12 recclap 9018 . . . . . . . 8 ((𝑄 ∈ ℂ ∧ 𝑄 # 0) → (1 / 𝑄) ∈ ℂ)
1312adantl 277 . . . . . . 7 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (1 / 𝑄) ∈ ℂ)
141, 2, 13mulassd 8356 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑄)) = (𝑃 · (𝑄 · (1 / 𝑄))))
15 recidap 9025 . . . . . . . 8 ((𝑄 ∈ ℂ ∧ 𝑄 # 0) → (𝑄 · (1 / 𝑄)) = 1)
1615oveq2d 6104 . . . . . . 7 ((𝑄 ∈ ℂ ∧ 𝑄 # 0) → (𝑃 · (𝑄 · (1 / 𝑄))) = (𝑃 · 1))
1716adantl 277 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · (𝑄 · (1 / 𝑄))) = (𝑃 · 1))
18 mulrid 8330 . . . . . . 7 (𝑃 ∈ ℂ → (𝑃 · 1) = 𝑃)
1918ad2antrr 492 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · 1) = 𝑃)
2014, 17, 193eqtrd 2275 . . . . 5 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑄)) = 𝑃)
2111, 20oveq12d 6106 . . . 4 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (((𝑃 · 𝑄) · (1 / 𝑃)) + ((𝑃 · 𝑄) · (1 / 𝑄))) = (𝑄 + 𝑃))
22 mulcl 8313 . . . . . 6 ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → (𝑃 · 𝑄) ∈ ℂ)
2322ad2ant2r 513 . . . . 5 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · 𝑄) ∈ ℂ)
2423, 4, 13adddid 8357 . . . 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 8471 . . . . 5 ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → (𝑃 + 𝑄) = (𝑄 + 𝑃))
2625ad2ant2r 513 . . . 4 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 + 𝑄) = (𝑄 + 𝑃))
2721, 24, 263eqtr4d 2281 . . 3 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = (𝑃 + 𝑄))
2822mulridd 8350 . . . 4 ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 · 𝑄) · 1) = (𝑃 · 𝑄))
2928ad2ant2r 513 . . 3 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · 1) = (𝑃 · 𝑄))
3027, 29eqeq12d 2253 . 2 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = ((𝑃 · 𝑄) · 1) ↔ (𝑃 + 𝑄) = (𝑃 · 𝑄)))
31 addcl 8311 . . . 4 (((1 / 𝑃) ∈ ℂ ∧ (1 / 𝑄) ∈ ℂ) → ((1 / 𝑃) + (1 / 𝑄)) ∈ ℂ)
323, 12, 31syl2an 289 . . 3 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((1 / 𝑃) + (1 / 𝑄)) ∈ ℂ)
33 mulap0 8991 . . 3 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · 𝑄) # 0)
34 ax-1cn 8279 . . . 4 1 ∈ ℂ
35 mulcanap 9002 . . . 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))
3634, 35mp3an2 1366 . . 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))
3732, 23, 33, 36syl12anc 1276 . 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 2240 . . . 4 ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ (𝑃 · 𝑄) = (𝑃 + 𝑄))
39 muleqadd 9007 . . . 4 ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 · 𝑄) = (𝑃 + 𝑄) ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1))
4038, 39bitrid 192 . . 3 ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1))
4140ad2ant2r 513 . 2 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1))
4230, 37, 413bitr3d 218 1 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (((1 / 𝑃) + (1 / 𝑄)) = 1 ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1))
Colors of variables:    wff set class
This proof depends on syntax axioms:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1402   ∈ wcel 2209   class class class wbr 4131  (class class class)co 6088  ℂcc 8184  0cc0 8186  1c1 8187   + caddc 8189   · cmul 8191   − cmin 8505   # cap 8918   / cdiv 9011
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-14 2212  ax-ext 2220  ax-sep 4251  ax-pow 4313  ax-pr 4348  ax-un 4580  ax-setind 4686  ax-cnex 8277  ax-resscn 8278  ax-1cn 8279  ax-1re 8280  ax-icn 8281  ax-addcl 8282  ax-addrcl 8283  ax-mulcl 8284  ax-mulrcl 8285  ax-addcom 8286  ax-mulcom 8287  ax-addass 8288  ax-mulass 8289  ax-distr 8290  ax-i2m1 8291  ax-0lt1 8292  ax-1rid 8293  ax-0id 8294  ax-rnegex 8295  ax-precex 8296  ax-cnre 8297  ax-pre-ltirr 8298  ax-pre-ltwlin 8299  ax-pre-lttrn 8300  ax-pre-apti 8301  ax-pre-ltadd 8302  ax-pre-mulgt0 8303  ax-pre-mulext 8304
This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-fal 1408  df-nf 1514  df-sb 1816  df-eu 2089  df-mo 2090  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ne 2421  df-nel 2516  df-ral 2533  df-rex 2534  df-reu 2535  df-rmo 2536  df-rab 2537  df-v 2823  df-sbc 3052  df-dif 3222  df-un 3224  df-in 3226  df-ss 3233  df-pw 3690  df-sn 3715  df-pr 3716  df-op 3718  df-uni 3936  df-br 4132  df-opab 4194  df-id 4440  df-po 4443  df-iso 4444  df-xp 4782  df-rel 4783  df-cnv 4784  df-co 4785  df-dm 4786  df-iota 5340  df-fun 5382  df-fv 5388  df-riota 6041  df-ov 6091  df-oprab 6092  df-mpo 6093  df-pnf 8369  df-mnf 8370  df-xr 8371  df-ltxr 8372  df-le 8373  df-sub 8507  df-neg 8508  df-reap 8912  df-ap 8919  df-div 9012
This theorem is used by: (None)
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