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Theorem conjmulap 8914
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 527 . . . . . . 7 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → 𝑃 ∈ ℂ)
2 simprl 531 . . . . . . 7 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → 𝑄 ∈ ℂ)
3 recclap 8864 . . . . . . . 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 8337 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑃)) = ((𝑃 · (1 / 𝑃)) · 𝑄))
6 recidap 8871 . . . . . . . 8 ((𝑃 ∈ ℂ ∧ 𝑃 # 0) → (𝑃 · (1 / 𝑃)) = 1)
76oveq1d 6038 . . . . . . 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 8182 . . . . . . 7 (𝑄 ∈ ℂ → (1 · 𝑄) = 𝑄)
109ad2antrl 490 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (1 · 𝑄) = 𝑄)
115, 8, 103eqtrd 2267 . . . . 5 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑃)) = 𝑄)
12 recclap 8864 . . . . . . . 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 8208 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑄)) = (𝑃 · (𝑄 · (1 / 𝑄))))
15 recidap 8871 . . . . . . . 8 ((𝑄 ∈ ℂ ∧ 𝑄 # 0) → (𝑄 · (1 / 𝑄)) = 1)
1615oveq2d 6039 . . . . . . 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 8181 . . . . . . 7 (𝑃 ∈ ℂ → (𝑃 · 1) = 𝑃)
1918ad2antrr 488 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · 1) = 𝑃)
2014, 17, 193eqtrd 2267 . . . . 5 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑄)) = 𝑃)
2111, 20oveq12d 6041 . . . 4 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (((𝑃 · 𝑄) · (1 / 𝑃)) + ((𝑃 · 𝑄) · (1 / 𝑄))) = (𝑄 + 𝑃))
22 mulcl 8164 . . . . . 6 ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → (𝑃 · 𝑄) ∈ ℂ)
2322ad2ant2r 509 . . . . 5 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · 𝑄) ∈ ℂ)
2423, 4, 13adddid 8209 . . . 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 8321 . . . . 5 ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → (𝑃 + 𝑄) = (𝑄 + 𝑃))
2625ad2ant2r 509 . . . 4 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 + 𝑄) = (𝑄 + 𝑃))
2721, 24, 263eqtr4d 2273 . . 3 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = (𝑃 + 𝑄))
2822mulridd 8201 . . . 4 ((𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ) → ((𝑃 · 𝑄) · 1) = (𝑃 · 𝑄))
2928ad2ant2r 509 . . 3 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · 1) = (𝑃 · 𝑄))
3027, 29eqeq12d 2245 . 2 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = ((𝑃 · 𝑄) · 1) ↔ (𝑃 + 𝑄) = (𝑃 · 𝑄)))
31 addcl 8162 . . . 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 8839 . . 3 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · 𝑄) # 0)
34 ax-1cn 8130 . . . 4 1 ∈ ℂ
35 mulcanap 8850 . . . 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 1361 . . 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 1271 . 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 2232 . . . 4 ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ (𝑃 · 𝑄) = (𝑃 + 𝑄))
39 muleqadd 8853 . . . 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 509 . 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
Syntax hints:  wi 4  wa 104  wb 105   = wceq 1397  wcel 2201   class class class wbr 4089  (class class class)co 6023  cc 8035  0cc0 8037  1c1 8038   + caddc 8040   · cmul 8042  cmin 8355   # cap 8766   / cdiv 8857
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 619  ax-in2 620  ax-io 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-13 2203  ax-14 2204  ax-ext 2212  ax-sep 4208  ax-pow 4266  ax-pr 4301  ax-un 4532  ax-setind 4637  ax-cnex 8128  ax-resscn 8129  ax-1cn 8130  ax-1re 8131  ax-icn 8132  ax-addcl 8133  ax-addrcl 8134  ax-mulcl 8135  ax-mulrcl 8136  ax-addcom 8137  ax-mulcom 8138  ax-addass 8139  ax-mulass 8140  ax-distr 8141  ax-i2m1 8142  ax-0lt1 8143  ax-1rid 8144  ax-0id 8145  ax-rnegex 8146  ax-precex 8147  ax-cnre 8148  ax-pre-ltirr 8149  ax-pre-ltwlin 8150  ax-pre-lttrn 8151  ax-pre-apti 8152  ax-pre-ltadd 8153  ax-pre-mulgt0 8154  ax-pre-mulext 8155
This theorem depends on definitions:  df-bi 117  df-3an 1006  df-tru 1400  df-fal 1403  df-nf 1509  df-sb 1810  df-eu 2081  df-mo 2082  df-clab 2217  df-cleq 2223  df-clel 2226  df-nfc 2362  df-ne 2402  df-nel 2497  df-ral 2514  df-rex 2515  df-reu 2516  df-rmo 2517  df-rab 2518  df-v 2803  df-sbc 3031  df-dif 3201  df-un 3203  df-in 3205  df-ss 3212  df-pw 3655  df-sn 3676  df-pr 3677  df-op 3679  df-uni 3895  df-br 4090  df-opab 4152  df-id 4392  df-po 4395  df-iso 4396  df-xp 4733  df-rel 4734  df-cnv 4735  df-co 4736  df-dm 4737  df-iota 5288  df-fun 5330  df-fv 5336  df-riota 5976  df-ov 6026  df-oprab 6027  df-mpo 6028  df-pnf 8221  df-mnf 8222  df-xr 8223  df-ltxr 8224  df-le 8225  df-sub 8357  df-neg 8358  df-reap 8760  df-ap 8767  df-div 8858
This theorem is referenced by: (None)
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