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Theorem addcan2 7966
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 7963 . . 3 (𝐶 ∈ ℂ → ∃𝑥 ∈ ℂ (𝐶 + 𝑥) = 0)
213ad2ant3 1005 . 2 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ∃𝑥 ∈ ℂ (𝐶 + 𝑥) = 0)
3 oveq1 5788 . . . 4 ((𝐴 + 𝐶) = (𝐵 + 𝐶) → ((𝐴 + 𝐶) + 𝑥) = ((𝐵 + 𝐶) + 𝑥))
4 simpl1 985 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝐴 ∈ ℂ)
5 simpl3 987 . . . . . . 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 7811 . . . . . 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 5797 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐴 + (𝐶 + 𝑥)) = (𝐴 + 0))
10 addid1 7923 . . . . . . 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 2177 . . . . 5 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐴 + 𝐶) + 𝑥) = 𝐴)
13 simpl2 986 . . . . . . 7 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝐵 ∈ ℂ)
1413, 5, 6addassd 7811 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐵 + 𝐶) + 𝑥) = (𝐵 + (𝐶 + 𝑥)))
158oveq2d 5797 . . . . . 6 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐵 + (𝐶 + 𝑥)) = (𝐵 + 0))
16 addid1 7923 . . . . . . 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 2177 . . . . 5 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐵 + 𝐶) + 𝑥) = 𝐵)
1912, 18eqeq12d 2155 . . . 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 5788 . . 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 2557 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 963   = wceq 1332  wcel 1481  wrex 2418  (class class class)co 5781  cc 7641  0cc0 7643   + caddc 7646
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 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122  ax-resscn 7735  ax-1cn 7736  ax-icn 7738  ax-addcl 7739  ax-addrcl 7740  ax-mulcl 7741  ax-addcom 7743  ax-addass 7745  ax-distr 7747  ax-i2m1 7748  ax-0id 7751  ax-rnegex 7752  ax-cnre 7754
This theorem depends on definitions:  df-bi 116  df-3an 965  df-tru 1335  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-ral 2422  df-rex 2423  df-v 2691  df-un 3079  df-in 3081  df-ss 3088  df-sn 3537  df-pr 3538  df-op 3540  df-uni 3744  df-br 3937  df-iota 5095  df-fv 5138  df-ov 5784
This theorem is referenced by:  addcan2i  7968  addcan2d  7970  muleqadd  8452
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