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Metamath Proof Explorer

Theorem renegadd

Description: Relationship between real negation and addition. (Contributed by Steven Nguyen, 7-Jan-2023)

Ref Expression
Assertion renegadd ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( 0 −ℝ 𝐴 ) = 𝐵 ↔ ( 𝐴 + 𝐵 ) = 0 ) )

Proof

Step Hyp Ref Expression
1 elre0re ⊢ ( 𝐴 ∈ ℝ → 0 ∈ ℝ )
2 resubval ⊢ ( ( 0 ∈ ℝ ∧ 𝐴 ∈ ℝ ) → ( 0 −ℝ 𝐴 ) = ( ℩ 𝑥 ∈ ℝ ( 𝐴 + 𝑥 ) = 0 ) )
3 1 2 mpancom ⊢ ( 𝐴 ∈ ℝ → ( 0 −ℝ 𝐴 ) = ( ℩ 𝑥 ∈ ℝ ( 𝐴 + 𝑥 ) = 0 ) )
4 3 eqeq1d ⊢ ( 𝐴 ∈ ℝ → ( ( 0 −ℝ 𝐴 ) = 𝐵 ↔ ( ℩ 𝑥 ∈ ℝ ( 𝐴 + 𝑥 ) = 0 ) = 𝐵 ) )
5 4 adantr ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( 0 −ℝ 𝐴 ) = 𝐵 ↔ ( ℩ 𝑥 ∈ ℝ ( 𝐴 + 𝑥 ) = 0 ) = 𝐵 ) )
6 renegeu ⊢ ( 𝐴 ∈ ℝ → ∃! 𝑥 ∈ ℝ ( 𝐴 + 𝑥 ) = 0 )
7 oveq2 ⊢ ( 𝑥 = 𝐵 → ( 𝐴 + 𝑥 ) = ( 𝐴 + 𝐵 ) )
8 7 eqeq1d ⊢ ( 𝑥 = 𝐵 → ( ( 𝐴 + 𝑥 ) = 0 ↔ ( 𝐴 + 𝐵 ) = 0 ) )
9 8 riota2 ⊢ ( ( 𝐵 ∈ ℝ ∧ ∃! 𝑥 ∈ ℝ ( 𝐴 + 𝑥 ) = 0 ) → ( ( 𝐴 + 𝐵 ) = 0 ↔ ( ℩ 𝑥 ∈ ℝ ( 𝐴 + 𝑥 ) = 0 ) = 𝐵 ) )
10 6 9 sylan2 ⊢ ( ( 𝐵 ∈ ℝ ∧ 𝐴 ∈ ℝ ) → ( ( 𝐴 + 𝐵 ) = 0 ↔ ( ℩ 𝑥 ∈ ℝ ( 𝐴 + 𝑥 ) = 0 ) = 𝐵 ) )
11 10 ancoms ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( 𝐴 + 𝐵 ) = 0 ↔ ( ℩ 𝑥 ∈ ℝ ( 𝐴 + 𝑥 ) = 0 ) = 𝐵 ) )
12 5 11 bitr4d ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( 0 −ℝ 𝐴 ) = 𝐵 ↔ ( 𝐴 + 𝐵 ) = 0 ) )