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Theorem addcan2 8079
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 8076 . . 3 (𝐶 ∈ ℂ → ∃𝑥 ∈ ℂ (𝐶 + 𝑥) = 0)
213ad2ant3 1010 . 2 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ∃𝑥 ∈ ℂ (𝐶 + 𝑥) = 0)
3 oveq1 5849 . . . 4 ((𝐴 + 𝐶) = (𝐵 + 𝐶) → ((𝐴 + 𝐶) + 𝑥) = ((𝐵 + 𝐶) + 𝑥))
4 simpl1 990 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝐴 ∈ ℂ)
5 simpl3 992 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝐶 ∈ ℂ)
6 simprl 521 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝑥 ∈ ℂ)
74, 5, 6addassd 7921 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐴 + 𝐶) + 𝑥) = (𝐴 + (𝐶 + 𝑥)))
8 simprr 522 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐶 + 𝑥) = 0)
98oveq2d 5858 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐴 + (𝐶 + 𝑥)) = (𝐴 + 0))
10 addid1 8036 . . . . . . 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 2202 . . . . 5 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐴 + 𝐶) + 𝑥) = 𝐴)
13 simpl2 991 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝐵 ∈ ℂ)
1413, 5, 6addassd 7921 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐵 + 𝐶) + 𝑥) = (𝐵 + (𝐶 + 𝑥)))
158oveq2d 5858 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐵 + (𝐶 + 𝑥)) = (𝐵 + 0))
16 addid1 8036 . . . . . . 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 2202 . . . . 5 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐵 + 𝐶) + 𝑥) = 𝐵)
1912, 18eqeq12d 2180 . . . 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 5849 . . 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 2588 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 968   = wceq 1343  wcel 2136  wrex 2445  (class class class)co 5842  cc 7751  0cc0 7753   + caddc 7756
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 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147  ax-resscn 7845  ax-1cn 7846  ax-icn 7848  ax-addcl 7849  ax-addrcl 7850  ax-mulcl 7851  ax-addcom 7853  ax-addass 7855  ax-distr 7857  ax-i2m1 7858  ax-0id 7861  ax-rnegex 7862  ax-cnre 7864
This theorem depends on definitions:  df-bi 116  df-3an 970  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-ral 2449  df-rex 2450  df-v 2728  df-un 3120  df-in 3122  df-ss 3129  df-sn 3582  df-pr 3583  df-op 3585  df-uni 3790  df-br 3983  df-iota 5153  df-fv 5196  df-ov 5845
This theorem is referenced by:  addcan2i  8081  addcan2d  8083  muleqadd  8565
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