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 Description: Addition commutes. This used to be one of our complex number axioms, until it was found to be dependent on the others. Based on ideas by Eric Schmidt. (Contributed by Scott Fenton, 3-Jan-2013.)
Assertion
Ref Expression
addcom ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴 + 𝐵) = (𝐵 + 𝐴))

StepHypRef Expression
1 1cnd 9935 . . . . . . . 8 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → 1 ∈ ℂ)
21, 1addcld 9938 . . . . . . 7 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (1 + 1) ∈ ℂ)
3 simpl 472 . . . . . . 7 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → 𝐴 ∈ ℂ)
4 simpr 476 . . . . . . 7 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → 𝐵 ∈ ℂ)
52, 3, 4adddid 9943 . . . . . 6 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((1 + 1) · (𝐴 + 𝐵)) = (((1 + 1) · 𝐴) + ((1 + 1) · 𝐵)))
63, 4addcld 9938 . . . . . . 7 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴 + 𝐵) ∈ ℂ)
7 1p1times 10086 . . . . . . 7 ((𝐴 + 𝐵) ∈ ℂ → ((1 + 1) · (𝐴 + 𝐵)) = ((𝐴 + 𝐵) + (𝐴 + 𝐵)))
86, 7syl 17 . . . . . 6 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((1 + 1) · (𝐴 + 𝐵)) = ((𝐴 + 𝐵) + (𝐴 + 𝐵)))
9 1p1times 10086 . . . . . . 7 (𝐴 ∈ ℂ → ((1 + 1) · 𝐴) = (𝐴 + 𝐴))
10 1p1times 10086 . . . . . . 7 (𝐵 ∈ ℂ → ((1 + 1) · 𝐵) = (𝐵 + 𝐵))
119, 10oveqan12d 6568 . . . . . 6 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (((1 + 1) · 𝐴) + ((1 + 1) · 𝐵)) = ((𝐴 + 𝐴) + (𝐵 + 𝐵)))
125, 8, 113eqtr3rd 2653 . . . . 5 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐴 + 𝐴) + (𝐵 + 𝐵)) = ((𝐴 + 𝐵) + (𝐴 + 𝐵)))
133, 3addcld 9938 . . . . . 6 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴 + 𝐴) ∈ ℂ)
1413, 4, 4addassd 9941 . . . . 5 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (((𝐴 + 𝐴) + 𝐵) + 𝐵) = ((𝐴 + 𝐴) + (𝐵 + 𝐵)))
156, 3, 4addassd 9941 . . . . 5 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (((𝐴 + 𝐵) + 𝐴) + 𝐵) = ((𝐴 + 𝐵) + (𝐴 + 𝐵)))
1612, 14, 153eqtr4d 2654 . . . 4 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (((𝐴 + 𝐴) + 𝐵) + 𝐵) = (((𝐴 + 𝐵) + 𝐴) + 𝐵))
1713, 4addcld 9938 . . . . 5 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐴 + 𝐴) + 𝐵) ∈ ℂ)
186, 3addcld 9938 . . . . 5 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐴 + 𝐵) + 𝐴) ∈ ℂ)
19 addcan2 10100 . . . . 5 ((((𝐴 + 𝐴) + 𝐵) ∈ ℂ ∧ ((𝐴 + 𝐵) + 𝐴) ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((((𝐴 + 𝐴) + 𝐵) + 𝐵) = (((𝐴 + 𝐵) + 𝐴) + 𝐵) ↔ ((𝐴 + 𝐴) + 𝐵) = ((𝐴 + 𝐵) + 𝐴)))
2017, 18, 4, 19syl3anc 1318 . . . 4 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((((𝐴 + 𝐴) + 𝐵) + 𝐵) = (((𝐴 + 𝐵) + 𝐴) + 𝐵) ↔ ((𝐴 + 𝐴) + 𝐵) = ((𝐴 + 𝐵) + 𝐴)))
2116, 20mpbid 221 . . 3 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐴 + 𝐴) + 𝐵) = ((𝐴 + 𝐵) + 𝐴))
223, 3, 4addassd 9941 . . 3 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐴 + 𝐴) + 𝐵) = (𝐴 + (𝐴 + 𝐵)))
233, 4, 3addassd 9941 . . 3 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐴 + 𝐵) + 𝐴) = (𝐴 + (𝐵 + 𝐴)))
2421, 22, 233eqtr3d 2652 . 2 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴 + (𝐴 + 𝐵)) = (𝐴 + (𝐵 + 𝐴)))
254, 3addcld 9938 . . 3 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐵 + 𝐴) ∈ ℂ)
26 addcan 10099 . . 3 ((𝐴 ∈ ℂ ∧ (𝐴 + 𝐵) ∈ ℂ ∧ (𝐵 + 𝐴) ∈ ℂ) → ((𝐴 + (𝐴 + 𝐵)) = (𝐴 + (𝐵 + 𝐴)) ↔ (𝐴 + 𝐵) = (𝐵 + 𝐴)))
273, 6, 25, 26syl3anc 1318 . 2 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐴 + (𝐴 + 𝐵)) = (𝐴 + (𝐵 + 𝐴)) ↔ (𝐴 + 𝐵) = (𝐵 + 𝐴)))
2824, 27mpbid 221 1 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴 + 𝐵) = (𝐵 + 𝐴))