Metamath Proof Explorer


Theorem subadd

Description: Relationship between subtraction and addition. (Contributed by NM, 20-Jan-1997) (Revised by Mario Carneiro, 21-Dec-2013)

Ref Expression
Assertion subadd ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐵 ) = 𝐶 ↔ ( 𝐵 + 𝐶 ) = 𝐴 ) )

Proof

Step Hyp Ref Expression
1 subval ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴𝐵 ) = ( 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) )
2 1 eqeq1d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴𝐵 ) = 𝐶 ↔ ( 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) = 𝐶 ) )
3 2 3adant3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐵 ) = 𝐶 ↔ ( 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) = 𝐶 ) )
4 negeu ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ∃! 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 )
5 oveq2 ( 𝑥 = 𝐶 → ( 𝐵 + 𝑥 ) = ( 𝐵 + 𝐶 ) )
6 5 eqeq1d ( 𝑥 = 𝐶 → ( ( 𝐵 + 𝑥 ) = 𝐴 ↔ ( 𝐵 + 𝐶 ) = 𝐴 ) )
7 6 riota2 ( ( 𝐶 ∈ ℂ ∧ ∃! 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) → ( ( 𝐵 + 𝐶 ) = 𝐴 ↔ ( 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) = 𝐶 ) )
8 4 7 sylan2 ( ( 𝐶 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) ) → ( ( 𝐵 + 𝐶 ) = 𝐴 ↔ ( 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) = 𝐶 ) )
9 8 3impb ( ( 𝐶 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( ( 𝐵 + 𝐶 ) = 𝐴 ↔ ( 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) = 𝐶 ) )
10 9 3com13 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 + 𝐶 ) = 𝐴 ↔ ( 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) = 𝐶 ) )
11 3 10 bitr4d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐵 ) = 𝐶 ↔ ( 𝐵 + 𝐶 ) = 𝐴 ) )