Theorem addcan2 8360

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.

Step	Hyp	Ref	Expression
1		cnegex 8357	. . 3 ⊢ (𝐶 ∈ ℂ → ∃𝑥 ∈ ℂ (𝐶 + 𝑥) = 0)
2	1	3ad2ant3 1046	. 2 ⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ∃𝑥 ∈ ℂ (𝐶 + 𝑥) = 0)
3		oveq1 6025	. . . 4 ⊢ ((𝐴 + 𝐶) = (𝐵 + 𝐶) → ((𝐴 + 𝐶) + 𝑥) = ((𝐵 + 𝐶) + 𝑥))
4		simpl1 1026	. . . . . . 7 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝐴 ∈ ℂ)
5		simpl3 1028	. . . . . . 7 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝐶 ∈ ℂ)
6		simprl 531	. . . . . . 7 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝑥 ∈ ℂ)
7	4, 5, 6	addassd 8202	. . . . . 6 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐴 + 𝐶) + 𝑥) = (𝐴 + (𝐶 + 𝑥)))
8		simprr 533	. . . . . . 7 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐶 + 𝑥) = 0)
9	8	oveq2d 6034	. . . . . 6 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐴 + (𝐶 + 𝑥)) = (𝐴 + 0))
10		addrid 8317	. . . . . . 7 ⊢ (𝐴 ∈ ℂ → (𝐴 + 0) = 𝐴)
11	4, 10	syl 14	. . . . . 6 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐴 + 0) = 𝐴)
12	7, 9, 11	3eqtrd 2268	. . . . 5 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐴 + 𝐶) + 𝑥) = 𝐴)
13		simpl2 1027	. . . . . . 7 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → 𝐵 ∈ ℂ)
14	13, 5, 6	addassd 8202	. . . . . 6 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐵 + 𝐶) + 𝑥) = (𝐵 + (𝐶 + 𝑥)))
15	8	oveq2d 6034	. . . . . 6 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐵 + (𝐶 + 𝑥)) = (𝐵 + 0))
16		addrid 8317	. . . . . . 7 ⊢ (𝐵 ∈ ℂ → (𝐵 + 0) = 𝐵)
17	13, 16	syl 14	. . . . . 6 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (𝐵 + 0) = 𝐵)
18	14, 15, 17	3eqtrd 2268	. . . . 5 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐵 + 𝐶) + 𝑥) = 𝐵)
19	12, 18	eqeq12d 2246	. . . 4 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → (((𝐴 + 𝐶) + 𝑥) = ((𝐵 + 𝐶) + 𝑥) ↔ 𝐴 = 𝐵))
20	3, 19	imbitrid 154	. . 3 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐴 + 𝐶) = (𝐵 + 𝐶) → 𝐴 = 𝐵))
21		oveq1 6025	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 + 𝐶) = (𝐵 + 𝐶))
22	20, 21	impbid1 142	. 2 ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝑥 ∈ ℂ ∧ (𝐶 + 𝑥) = 0)) → ((𝐴 + 𝐶) = (𝐵 + 𝐶) ↔ 𝐴 = 𝐵))
23	2, 22	rexlimddv 2655	1 ⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ((𝐴 + 𝐶) = (𝐵 + 𝐶) ↔ 𝐴 = 𝐵))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   ∧ w3a 1004   = wceq 1397   ∈ wcel 2202  ∃wrex 2511  (class class class)co 6018  ℂcc 8030  0cc0 8032   + caddc 8035

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213  ax-resscn 8124  ax-1cn 8125  ax-icn 8127  ax-addcl 8128  ax-addrcl 8129  ax-mulcl 8130  ax-addcom 8132  ax-addass 8134  ax-distr 8136  ax-i2m1 8137  ax-0id 8140  ax-rnegex 8141  ax-cnre 8143

This theorem depends on definitions:  df-bi 117  df-3an 1006  df-tru 1400  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-ral 2515  df-rex 2516  df-v 2804  df-un 3204  df-in 3206  df-ss 3213  df-sn 3675  df-pr 3676  df-op 3678  df-uni 3894  df-br 4089  df-iota 5286  df-fv 5334  df-ov 6021

This theorem is referenced by:  addcan2i  8362  addcan2d  8364  muleqadd  8848