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Theorem conjmulap 8489
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 518 . . . . . . 7 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → 𝑃 ∈ ℂ)
2 simprl 520 . . . . . . 7 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → 𝑄 ∈ ℂ)
3 recclap 8439 . . . . . . . 8 ((𝑃 ∈ ℂ ∧ 𝑃 # 0) → (1 / 𝑃) ∈ ℂ)
43adantr 274 . . . . . . 7 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (1 / 𝑃) ∈ ℂ)
51, 2, 4mul32d 7915 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑃)) = ((𝑃 · (1 / 𝑃)) · 𝑄))
6 recidap 8446 . . . . . . . 8 ((𝑃 ∈ ℂ ∧ 𝑃 # 0) → (𝑃 · (1 / 𝑃)) = 1)
76oveq1d 5789 . . . . . . 7 ((𝑃 ∈ ℂ ∧ 𝑃 # 0) → ((𝑃 · (1 / 𝑃)) · 𝑄) = (1 · 𝑄))
87adantr 274 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · (1 / 𝑃)) · 𝑄) = (1 · 𝑄))
9 mulid2 7764 . . . . . . 7 (𝑄 ∈ ℂ → (1 · 𝑄) = 𝑄)
109ad2antrl 481 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (1 · 𝑄) = 𝑄)
115, 8, 103eqtrd 2176 . . . . 5 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑃)) = 𝑄)
12 recclap 8439 . . . . . . . 8 ((𝑄 ∈ ℂ ∧ 𝑄 # 0) → (1 / 𝑄) ∈ ℂ)
1312adantl 275 . . . . . . 7 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (1 / 𝑄) ∈ ℂ)
141, 2, 13mulassd 7789 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑄)) = (𝑃 · (𝑄 · (1 / 𝑄))))
15 recidap 8446 . . . . . . . 8 ((𝑄 ∈ ℂ ∧ 𝑄 # 0) → (𝑄 · (1 / 𝑄)) = 1)
1615oveq2d 5790 . . . . . . 7 ((𝑄 ∈ ℂ ∧ 𝑄 # 0) → (𝑃 · (𝑄 · (1 / 𝑄))) = (𝑃 · 1))
1716adantl 275 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · (𝑄 · (1 / 𝑄))) = (𝑃 · 1))
18 mulid1 7763 . . . . . . 7 (𝑃 ∈ ℂ → (𝑃 · 1) = 𝑃)
1918ad2antrr 479 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · 1) = 𝑃)
2014, 17, 193eqtrd 2176 . . . . 5 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑄)) = 𝑃)
2111, 20oveq12d 5792 . . . 4 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (((𝑃 · 𝑄) · (1 / 𝑃)) + ((𝑃 · 𝑄) · (1 / 𝑄))) = (𝑄 + 𝑃))
22 mulcl 7747 . . . . . 6 ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → (𝑃 · 𝑄) ∈ ℂ)
2322ad2ant2r 500 . . . . 5 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · 𝑄) ∈ ℂ)
2423, 4, 13adddid 7790 . . . 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 7899 . . . . 5 ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → (𝑃 + 𝑄) = (𝑄 + 𝑃))
2625ad2ant2r 500 . . . 4 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 + 𝑄) = (𝑄 + 𝑃))
2721, 24, 263eqtr4d 2182 . . 3 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = (𝑃 + 𝑄))
2822mulid1d 7783 . . . 4 ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 · 𝑄) · 1) = (𝑃 · 𝑄))
2928ad2ant2r 500 . . 3 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · 1) = (𝑃 · 𝑄))
3027, 29eqeq12d 2154 . 2 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = ((𝑃 · 𝑄) · 1) ↔ (𝑃 + 𝑄) = (𝑃 · 𝑄)))
31 addcl 7745 . . . 4 (((1 / 𝑃) ∈ ℂ ∧ (1 / 𝑄) ∈ ℂ) → ((1 / 𝑃) + (1 / 𝑄)) ∈ ℂ)
323, 12, 31syl2an 287 . . 3 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((1 / 𝑃) + (1 / 𝑄)) ∈ ℂ)
33 mulap0 8415 . . 3 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · 𝑄) # 0)
34 ax-1cn 7713 . . . 4 1 ∈ ℂ
35 mulcanap 8426 . . . 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 1303 . . 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 1214 . 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 2141 . . . 4 ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ (𝑃 · 𝑄) = (𝑃 + 𝑄))
39 muleqadd 8429 . . . 4 ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 · 𝑄) = (𝑃 + 𝑄) ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1))
4038, 39syl5bb 191 . . 3 ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1))
4140ad2ant2r 500 . 2 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ ((𝑃 − 1) · (𝑄 − 1)) = 1))
4230, 37, 413bitr3d 217 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
Syntax hints:  wi 4  wa 103  wb 104   = wceq 1331  wcel 1480   class class class wbr 3929  (class class class)co 5774  cc 7618  0cc0 7620  1c1 7621   + caddc 7623   · cmul 7625  cmin 7933   # cap 8343   / cdiv 8432
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 603  ax-in2 604  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-13 1491  ax-14 1492  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121  ax-sep 4046  ax-pow 4098  ax-pr 4131  ax-un 4355  ax-setind 4452  ax-cnex 7711  ax-resscn 7712  ax-1cn 7713  ax-1re 7714  ax-icn 7715  ax-addcl 7716  ax-addrcl 7717  ax-mulcl 7718  ax-mulrcl 7719  ax-addcom 7720  ax-mulcom 7721  ax-addass 7722  ax-mulass 7723  ax-distr 7724  ax-i2m1 7725  ax-0lt1 7726  ax-1rid 7727  ax-0id 7728  ax-rnegex 7729  ax-precex 7730  ax-cnre 7731  ax-pre-ltirr 7732  ax-pre-ltwlin 7733  ax-pre-lttrn 7734  ax-pre-apti 7735  ax-pre-ltadd 7736  ax-pre-mulgt0 7737  ax-pre-mulext 7738
This theorem depends on definitions:  df-bi 116  df-3an 964  df-tru 1334  df-fal 1337  df-nf 1437  df-sb 1736  df-eu 2002  df-mo 2003  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-ne 2309  df-nel 2404  df-ral 2421  df-rex 2422  df-reu 2423  df-rmo 2424  df-rab 2425  df-v 2688  df-sbc 2910  df-dif 3073  df-un 3075  df-in 3077  df-ss 3084  df-pw 3512  df-sn 3533  df-pr 3534  df-op 3536  df-uni 3737  df-br 3930  df-opab 3990  df-id 4215  df-po 4218  df-iso 4219  df-xp 4545  df-rel 4546  df-cnv 4547  df-co 4548  df-dm 4549  df-iota 5088  df-fun 5125  df-fv 5131  df-riota 5730  df-ov 5777  df-oprab 5778  df-mpo 5779  df-pnf 7802  df-mnf 7803  df-xr 7804  df-ltxr 7805  df-le 7806  df-sub 7935  df-neg 7936  df-reap 8337  df-ap 8344  df-div 8433
This theorem is referenced by: (None)
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