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Theorem addcan2 7911
Description: Cancellation law for addition. (Contributed by NM, 30-Jul-2004.) (Revised by Scott Fenton, 3-Jan-2013.)
Assertion
Ref Expression
addcan2 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ((𝐴 + 𝐶) = (𝐵 + 𝐶) ↔ 𝐴 = 𝐵))

Proof of Theorem addcan2
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 cnegex 7908 . . 3 (𝐶 ∈ ℂ → ∃𝑥 ∈ ℂ (𝐶 + 𝑥) = 0)
213ad2ant3 989 . 2 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ∃𝑥 ∈ ℂ (𝐶 + 𝑥) = 0)
3 oveq1 5749 . . . 4 ((𝐴 + 𝐶) = (𝐵 + 𝐶) → ((𝐴 + 𝐶) + 𝑥) = ((𝐵 + 𝐶) + 𝑥))
4 simpl1 969 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝐴 ∈ ℂ)
5 simpl3 971 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝐶 ∈ ℂ)
6 simprl 505 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝑥 ∈ ℂ)
74, 5, 6addassd 7756 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐴 + 𝐶) + 𝑥) = (𝐴 + (𝐶 + 𝑥)))
8 simprr 506 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐶 + 𝑥) = 0)
98oveq2d 5758 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐴 + (𝐶 + 𝑥)) = (𝐴 + 0))
10 addid1 7868 . . . . . . 7 (𝐴 ∈ ℂ → (𝐴 + 0) = 𝐴)
114, 10syl 14 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐴 + 0) = 𝐴)
127, 9, 113eqtrd 2154 . . . . 5 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐴 + 𝐶) + 𝑥) = 𝐴)
13 simpl2 970 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝐵 ∈ ℂ)
1413, 5, 6addassd 7756 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐵 + 𝐶) + 𝑥) = (𝐵 + (𝐶 + 𝑥)))
158oveq2d 5758 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐵 + (𝐶 + 𝑥)) = (𝐵 + 0))
16 addid1 7868 . . . . . . 7 (𝐵 ∈ ℂ → (𝐵 + 0) = 𝐵)
1713, 16syl 14 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐵 + 0) = 𝐵)
1814, 15, 173eqtrd 2154 . . . . 5 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐵 + 𝐶) + 𝑥) = 𝐵)
1912, 18eqeq12d 2132 . . . 4 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (((𝐴 + 𝐶) + 𝑥) = ((𝐵 + 𝐶) + 𝑥) ↔ 𝐴 = 𝐵))
203, 19syl5ib 153 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐴 + 𝐶) = (𝐵 + 𝐶) → 𝐴 = 𝐵))
21 oveq1 5749 . . 3 (𝐴 = 𝐵 → (𝐴 + 𝐶) = (𝐵 + 𝐶))
2220, 21impbid1 141 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐴 + 𝐶) = (𝐵 + 𝐶) ↔ 𝐴 = 𝐵))
232, 22rexlimddv 2531 1 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ((𝐴 + 𝐶) = (𝐵 + 𝐶) ↔ 𝐴 = 𝐵))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103  wb 104  w3a 947   = wceq 1316  wcel 1465  wrex 2394  (class class class)co 5742  cc 7586  0cc0 7588   + caddc 7591
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-io 683  ax-5 1408  ax-7 1409  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-8 1467  ax-10 1468  ax-11 1469  ax-i12 1470  ax-bndl 1471  ax-4 1472  ax-17 1491  ax-i9 1495  ax-ial 1499  ax-i5r 1500  ax-ext 2099  ax-resscn 7680  ax-1cn 7681  ax-icn 7683  ax-addcl 7684  ax-addrcl 7685  ax-mulcl 7686  ax-addcom 7688  ax-addass 7690  ax-distr 7692  ax-i2m1 7693  ax-0id 7696  ax-rnegex 7697  ax-cnre 7699
This theorem depends on definitions:  df-bi 116  df-3an 949  df-tru 1319  df-nf 1422  df-sb 1721  df-clab 2104  df-cleq 2110  df-clel 2113  df-nfc 2247  df-ral 2398  df-rex 2399  df-v 2662  df-un 3045  df-in 3047  df-ss 3054  df-sn 3503  df-pr 3504  df-op 3506  df-uni 3707  df-br 3900  df-iota 5058  df-fv 5101  df-ov 5745
This theorem is referenced by:  addcan2i  7913  addcan2d  7915  muleqadd  8397
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