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Theorem conjmulap 8748
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 529 . . . . . . 7 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → 𝑄 ∈ ℂ)
3 recclap 8698 . . . . . . . 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 8172 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑃)) = ((𝑃 · (1 / 𝑃)) · 𝑄))
6 recidap 8705 . . . . . . . 8 ((𝑃 ∈ ℂ ∧ 𝑃 # 0) → (𝑃 · (1 / 𝑃)) = 1)
76oveq1d 5933 . . . . . . 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 8017 . . . . . . 7 (𝑄 ∈ ℂ → (1 · 𝑄) = 𝑄)
109ad2antrl 490 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (1 · 𝑄) = 𝑄)
115, 8, 103eqtrd 2230 . . . . 5 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑃)) = 𝑄)
12 recclap 8698 . . . . . . . 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 8043 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑄)) = (𝑃 · (𝑄 · (1 / 𝑄))))
15 recidap 8705 . . . . . . . 8 ((𝑄 ∈ ℂ ∧ 𝑄 # 0) → (𝑄 · (1 / 𝑄)) = 1)
1615oveq2d 5934 . . . . . . 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 8016 . . . . . . 7 (𝑃 ∈ ℂ → (𝑃 · 1) = 𝑃)
1918ad2antrr 488 . . . . . 6 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · 1) = 𝑃)
2014, 17, 193eqtrd 2230 . . . . 5 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · (1 / 𝑄)) = 𝑃)
2111, 20oveq12d 5936 . . . 4 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (((𝑃 · 𝑄) · (1 / 𝑃)) + ((𝑃 · 𝑄) · (1 / 𝑄))) = (𝑄 + 𝑃))
22 mulcl 7999 . . . . . 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 8044 . . . 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 8156 . . . . 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 2236 . . 3 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → ((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = (𝑃 + 𝑄))
2822mulridd 8036 . . . 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 2208 . 2 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (((𝑃 · 𝑄) · ((1 / 𝑃) + (1 / 𝑄))) = ((𝑃 · 𝑄) · 1) ↔ (𝑃 + 𝑄) = (𝑃 · 𝑄)))
31 addcl 7997 . . . 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 8673 . . 3 (((𝑃 ∈ ℂ ∧ 𝑃 # 0) ∧ (𝑄 ∈ ℂ ∧ 𝑄 # 0)) → (𝑃 · 𝑄) # 0)
34 ax-1cn 7965 . . . 4 1 ∈ ℂ
35 mulcanap 8684 . . . 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 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))
3732, 23, 33, 36syl12anc 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 2195 . . . 4 ((𝑃 + 𝑄) = (𝑃 · 𝑄) ↔ (𝑃 · 𝑄) = (𝑃 + 𝑄))
39 muleqadd 8687 . . . 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 1364  wcel 2164   class class class wbr 4029  (class class class)co 5918  cc 7870  0cc0 7872  1c1 7873   + caddc 7875   · cmul 7877  cmin 8190   # cap 8600   / cdiv 8691
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 615  ax-in2 616  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-13 2166  ax-14 2167  ax-ext 2175  ax-sep 4147  ax-pow 4203  ax-pr 4238  ax-un 4464  ax-setind 4569  ax-cnex 7963  ax-resscn 7964  ax-1cn 7965  ax-1re 7966  ax-icn 7967  ax-addcl 7968  ax-addrcl 7969  ax-mulcl 7970  ax-mulrcl 7971  ax-addcom 7972  ax-mulcom 7973  ax-addass 7974  ax-mulass 7975  ax-distr 7976  ax-i2m1 7977  ax-0lt1 7978  ax-1rid 7979  ax-0id 7980  ax-rnegex 7981  ax-precex 7982  ax-cnre 7983  ax-pre-ltirr 7984  ax-pre-ltwlin 7985  ax-pre-lttrn 7986  ax-pre-apti 7987  ax-pre-ltadd 7988  ax-pre-mulgt0 7989  ax-pre-mulext 7990
This theorem depends on definitions:  df-bi 117  df-3an 982  df-tru 1367  df-fal 1370  df-nf 1472  df-sb 1774  df-eu 2045  df-mo 2046  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-ne 2365  df-nel 2460  df-ral 2477  df-rex 2478  df-reu 2479  df-rmo 2480  df-rab 2481  df-v 2762  df-sbc 2986  df-dif 3155  df-un 3157  df-in 3159  df-ss 3166  df-pw 3603  df-sn 3624  df-pr 3625  df-op 3627  df-uni 3836  df-br 4030  df-opab 4091  df-id 4324  df-po 4327  df-iso 4328  df-xp 4665  df-rel 4666  df-cnv 4667  df-co 4668  df-dm 4669  df-iota 5215  df-fun 5256  df-fv 5262  df-riota 5873  df-ov 5921  df-oprab 5922  df-mpo 5923  df-pnf 8056  df-mnf 8057  df-xr 8058  df-ltxr 8059  df-le 8060  df-sub 8192  df-neg 8193  df-reap 8594  df-ap 8601  df-div 8692
This theorem is referenced by: (None)
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