Metamath Proof Explorer


Theorem addid0

Description: If adding a number to a another number yields the other number, the added number must be 0 . This shows that 0 is the unique (right) identity of the complex numbers. (Contributed by AV, 17-Jan-2021)

Ref Expression
Assertion addid0 ( ( 𝑋 ∈ ℂ ∧ 𝑌 ∈ ℂ ) → ( ( 𝑋 + 𝑌 ) = 𝑋𝑌 = 0 ) )

Proof

Step Hyp Ref Expression
1 simpl ( ( 𝑋 ∈ ℂ ∧ 𝑌 ∈ ℂ ) → 𝑋 ∈ ℂ )
2 simpr ( ( 𝑋 ∈ ℂ ∧ 𝑌 ∈ ℂ ) → 𝑌 ∈ ℂ )
3 1 1 2 subaddd ( ( 𝑋 ∈ ℂ ∧ 𝑌 ∈ ℂ ) → ( ( 𝑋𝑋 ) = 𝑌 ↔ ( 𝑋 + 𝑌 ) = 𝑋 ) )
4 eqcom ( ( 𝑋𝑋 ) = 𝑌𝑌 = ( 𝑋𝑋 ) )
5 simpr ( ( 𝑋 ∈ ℂ ∧ 𝑌 = ( 𝑋𝑋 ) ) → 𝑌 = ( 𝑋𝑋 ) )
6 subid ( 𝑋 ∈ ℂ → ( 𝑋𝑋 ) = 0 )
7 6 adantr ( ( 𝑋 ∈ ℂ ∧ 𝑌 = ( 𝑋𝑋 ) ) → ( 𝑋𝑋 ) = 0 )
8 5 7 eqtrd ( ( 𝑋 ∈ ℂ ∧ 𝑌 = ( 𝑋𝑋 ) ) → 𝑌 = 0 )
9 8 ex ( 𝑋 ∈ ℂ → ( 𝑌 = ( 𝑋𝑋 ) → 𝑌 = 0 ) )
10 4 9 syl5bi ( 𝑋 ∈ ℂ → ( ( 𝑋𝑋 ) = 𝑌𝑌 = 0 ) )
11 10 adantr ( ( 𝑋 ∈ ℂ ∧ 𝑌 ∈ ℂ ) → ( ( 𝑋𝑋 ) = 𝑌𝑌 = 0 ) )
12 3 11 sylbird ( ( 𝑋 ∈ ℂ ∧ 𝑌 ∈ ℂ ) → ( ( 𝑋 + 𝑌 ) = 𝑋𝑌 = 0 ) )
13 oveq2 ( 𝑌 = 0 → ( 𝑋 + 𝑌 ) = ( 𝑋 + 0 ) )
14 addid1 ( 𝑋 ∈ ℂ → ( 𝑋 + 0 ) = 𝑋 )
15 13 14 sylan9eqr ( ( 𝑋 ∈ ℂ ∧ 𝑌 = 0 ) → ( 𝑋 + 𝑌 ) = 𝑋 )
16 15 ex ( 𝑋 ∈ ℂ → ( 𝑌 = 0 → ( 𝑋 + 𝑌 ) = 𝑋 ) )
17 16 adantr ( ( 𝑋 ∈ ℂ ∧ 𝑌 ∈ ℂ ) → ( 𝑌 = 0 → ( 𝑋 + 𝑌 ) = 𝑋 ) )
18 12 17 impbid ( ( 𝑋 ∈ ℂ ∧ 𝑌 ∈ ℂ ) → ( ( 𝑋 + 𝑌 ) = 𝑋𝑌 = 0 ) )