ILE Home Intuitionistic Logic Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  ILE Home  >  Th. List  >  cnegexlem1 GIF version

Theorem cnegexlem1 7636
Description: Addition cancellation of a real number from two complex numbers. Lemma for cnegex 7639. (Contributed by Eric Schmidt, 22-May-2007.)
Assertion
Ref Expression
cnegexlem1 ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ((𝐴 + 𝐵) = (𝐴 + 𝐶) ↔ 𝐵 = 𝐶))

Proof of Theorem cnegexlem1
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 ax-rnegex 7433 . . . 4 (𝐴 ∈ ℝ → ∃𝑥 ∈ ℝ (𝐴 + 𝑥) = 0)
213ad2ant1 964 . . 3 ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ∃𝑥 ∈ ℝ (𝐴 + 𝑥) = 0)
3 recn 7454 . . . 4 (𝐴 ∈ ℝ → 𝐴 ∈ ℂ)
4 recn 7454 . . . . . . 7 (𝑥 ∈ ℝ → 𝑥 ∈ ℂ)
5 oveq2 5642 . . . . . . . . . . 11 ((𝐴 + 𝐵) = (𝐴 + 𝐶) → (𝑥 + (𝐴 + 𝐵)) = (𝑥 + (𝐴 + 𝐶)))
6 simpr 108 . . . . . . . . . . . . 13 (((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) → 𝑥 ∈ ℂ)
7 simpll 496 . . . . . . . . . . . . 13 (((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) → 𝐴 ∈ ℂ)
8 simplrl 502 . . . . . . . . . . . . 13 (((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) → 𝐵 ∈ ℂ)
96, 7, 8addassd 7489 . . . . . . . . . . . 12 (((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) → ((𝑥 + 𝐴) + 𝐵) = (𝑥 + (𝐴 + 𝐵)))
10 simplrr 503 . . . . . . . . . . . . 13 (((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) → 𝐶 ∈ ℂ)
116, 7, 10addassd 7489 . . . . . . . . . . . 12 (((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) → ((𝑥 + 𝐴) + 𝐶) = (𝑥 + (𝐴 + 𝐶)))
129, 11eqeq12d 2102 . . . . . . . . . . 11 (((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) → (((𝑥 + 𝐴) + 𝐵) = ((𝑥 + 𝐴) + 𝐶) ↔ (𝑥 + (𝐴 + 𝐵)) = (𝑥 + (𝐴 + 𝐶))))
135, 12syl5ibr 154 . . . . . . . . . 10 (((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) → ((𝐴 + 𝐵) = (𝐴 + 𝐶) → ((𝑥 + 𝐴) + 𝐵) = ((𝑥 + 𝐴) + 𝐶)))
1413adantr 270 . . . . . . . . 9 ((((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) ∧ (𝐴 + 𝑥) = 0) → ((𝐴 + 𝐵) = (𝐴 + 𝐶) → ((𝑥 + 𝐴) + 𝐵) = ((𝑥 + 𝐴) + 𝐶)))
15 addcom 7598 . . . . . . . . . . . . 13 ((𝐴 ∈ ℂ ∧ 𝑥 ∈ ℂ) → (𝐴 + 𝑥) = (𝑥 + 𝐴))
1615eqeq1d 2096 . . . . . . . . . . . 12 ((𝐴 ∈ ℂ ∧ 𝑥 ∈ ℂ) → ((𝐴 + 𝑥) = 0 ↔ (𝑥 + 𝐴) = 0))
1716adantlr 461 . . . . . . . . . . 11 (((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) → ((𝐴 + 𝑥) = 0 ↔ (𝑥 + 𝐴) = 0))
18 oveq1 5641 . . . . . . . . . . . . . . 15 ((𝑥 + 𝐴) = 0 → ((𝑥 + 𝐴) + 𝐵) = (0 + 𝐵))
19 oveq1 5641 . . . . . . . . . . . . . . 15 ((𝑥 + 𝐴) = 0 → ((𝑥 + 𝐴) + 𝐶) = (0 + 𝐶))
2018, 19eqeq12d 2102 . . . . . . . . . . . . . 14 ((𝑥 + 𝐴) = 0 → (((𝑥 + 𝐴) + 𝐵) = ((𝑥 + 𝐴) + 𝐶) ↔ (0 + 𝐵) = (0 + 𝐶)))
2120adantl 271 . . . . . . . . . . . . 13 ((((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) ∧ (𝑥 + 𝐴) = 0) → (((𝑥 + 𝐴) + 𝐵) = ((𝑥 + 𝐴) + 𝐶) ↔ (0 + 𝐵) = (0 + 𝐶)))
22 addid2 7600 . . . . . . . . . . . . . . . 16 (𝐵 ∈ ℂ → (0 + 𝐵) = 𝐵)
23 addid2 7600 . . . . . . . . . . . . . . . 16 (𝐶 ∈ ℂ → (0 + 𝐶) = 𝐶)
2422, 23eqeqan12d 2103 . . . . . . . . . . . . . . 15 ((𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ((0 + 𝐵) = (0 + 𝐶) ↔ 𝐵 = 𝐶))
2524adantl 271 . . . . . . . . . . . . . 14 ((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) → ((0 + 𝐵) = (0 + 𝐶) ↔ 𝐵 = 𝐶))
2625ad2antrr 472 . . . . . . . . . . . . 13 ((((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) ∧ (𝑥 + 𝐴) = 0) → ((0 + 𝐵) = (0 + 𝐶) ↔ 𝐵 = 𝐶))
2721, 26bitrd 186 . . . . . . . . . . . 12 ((((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) ∧ (𝑥 + 𝐴) = 0) → (((𝑥 + 𝐴) + 𝐵) = ((𝑥 + 𝐴) + 𝐶) ↔ 𝐵 = 𝐶))
2827ex 113 . . . . . . . . . . 11 (((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) → ((𝑥 + 𝐴) = 0 → (((𝑥 + 𝐴) + 𝐵) = ((𝑥 + 𝐴) + 𝐶) ↔ 𝐵 = 𝐶)))
2917, 28sylbid 148 . . . . . . . . . 10 (((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) → ((𝐴 + 𝑥) = 0 → (((𝑥 + 𝐴) + 𝐵) = ((𝑥 + 𝐴) + 𝐶) ↔ 𝐵 = 𝐶)))
3029imp 122 . . . . . . . . 9 ((((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) ∧ (𝐴 + 𝑥) = 0) → (((𝑥 + 𝐴) + 𝐵) = ((𝑥 + 𝐴) + 𝐶) ↔ 𝐵 = 𝐶))
3114, 30sylibd 147 . . . . . . . 8 ((((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) ∧ (𝐴 + 𝑥) = 0) → ((𝐴 + 𝐵) = (𝐴 + 𝐶) → 𝐵 = 𝐶))
3231ex 113 . . . . . . 7 (((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℂ) → ((𝐴 + 𝑥) = 0 → ((𝐴 + 𝐵) = (𝐴 + 𝐶) → 𝐵 = 𝐶)))
334, 32sylan2 280 . . . . . 6 (((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) ∧ 𝑥 ∈ ℝ) → ((𝐴 + 𝑥) = 0 → ((𝐴 + 𝐵) = (𝐴 + 𝐶) → 𝐵 = 𝐶)))
3433rexlimdva 2489 . . . . 5 ((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ)) → (∃𝑥 ∈ ℝ (𝐴 + 𝑥) = 0 → ((𝐴 + 𝐵) = (𝐴 + 𝐶) → 𝐵 = 𝐶)))
35343impb 1139 . . . 4 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (∃𝑥 ∈ ℝ (𝐴 + 𝑥) = 0 → ((𝐴 + 𝐵) = (𝐴 + 𝐶) → 𝐵 = 𝐶)))
363, 35syl3an1 1207 . . 3 ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (∃𝑥 ∈ ℝ (𝐴 + 𝑥) = 0 → ((𝐴 + 𝐵) = (𝐴 + 𝐶) → 𝐵 = 𝐶)))
372, 36mpd 13 . 2 ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ((𝐴 + 𝐵) = (𝐴 + 𝐶) → 𝐵 = 𝐶))
38 oveq2 5642 . 2 (𝐵 = 𝐶 → (𝐴 + 𝐵) = (𝐴 + 𝐶))
3937, 38impbid1 140 1 ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ((𝐴 + 𝐵) = (𝐴 + 𝐶) ↔ 𝐵 = 𝐶))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102  wb 103  w3a 924   = wceq 1289  wcel 1438  wrex 2360  (class class class)co 5634  cc 7327  cr 7328  0cc0 7329   + caddc 7332
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-bndl 1444  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473  ax-ext 2070  ax-resscn 7416  ax-1cn 7417  ax-icn 7419  ax-addcl 7420  ax-mulcl 7422  ax-addcom 7424  ax-addass 7426  ax-i2m1 7429  ax-0id 7432  ax-rnegex 7433
This theorem depends on definitions:  df-bi 115  df-3an 926  df-tru 1292  df-nf 1395  df-sb 1693  df-clab 2075  df-cleq 2081  df-clel 2084  df-nfc 2217  df-ral 2364  df-rex 2365  df-v 2621  df-un 3001  df-in 3003  df-ss 3010  df-sn 3447  df-pr 3448  df-op 3450  df-uni 3649  df-br 3838  df-iota 4967  df-fv 5010  df-ov 5637
This theorem is referenced by:  cnegexlem3  7638
  Copyright terms: Public domain W3C validator