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Theorem addcan2 8100
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 8097 . . 3 (𝐶 ∈ ℂ → ∃𝑥 ∈ ℂ (𝐶 + 𝑥) = 0)
213ad2ant3 1015 . 2 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ∃𝑥 ∈ ℂ (𝐶 + 𝑥) = 0)
3 oveq1 5860 . . . 4 ((𝐴 + 𝐶) = (𝐵 + 𝐶) → ((𝐴 + 𝐶) + 𝑥) = ((𝐵 + 𝐶) + 𝑥))
4 simpl1 995 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝐴 ∈ ℂ)
5 simpl3 997 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝐶 ∈ ℂ)
6 simprl 526 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝑥 ∈ ℂ)
74, 5, 6addassd 7942 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐴 + 𝐶) + 𝑥) = (𝐴 + (𝐶 + 𝑥)))
8 simprr 527 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐶 + 𝑥) = 0)
98oveq2d 5869 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐴 + (𝐶 + 𝑥)) = (𝐴 + 0))
10 addid1 8057 . . . . . . 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 2207 . . . . 5 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐴 + 𝐶) + 𝑥) = 𝐴)
13 simpl2 996 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝐵 ∈ ℂ)
1413, 5, 6addassd 7942 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐵 + 𝐶) + 𝑥) = (𝐵 + (𝐶 + 𝑥)))
158oveq2d 5869 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐵 + (𝐶 + 𝑥)) = (𝐵 + 0))
16 addid1 8057 . . . . . . 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 2207 . . . . 5 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐵 + 𝐶) + 𝑥) = 𝐵)
1912, 18eqeq12d 2185 . . . 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 5860 . . 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 2592 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 973   = wceq 1348  wcel 2141  wrex 2449  (class class class)co 5853  cc 7772  0cc0 7774   + caddc 7777
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 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152  ax-resscn 7866  ax-1cn 7867  ax-icn 7869  ax-addcl 7870  ax-addrcl 7871  ax-mulcl 7872  ax-addcom 7874  ax-addass 7876  ax-distr 7878  ax-i2m1 7879  ax-0id 7882  ax-rnegex 7883  ax-cnre 7885
This theorem depends on definitions:  df-bi 116  df-3an 975  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-ral 2453  df-rex 2454  df-v 2732  df-un 3125  df-in 3127  df-ss 3134  df-sn 3589  df-pr 3590  df-op 3592  df-uni 3797  df-br 3990  df-iota 5160  df-fv 5206  df-ov 5856
This theorem is referenced by:  addcan2i  8102  addcan2d  8104  muleqadd  8586
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