Metamath Proof Explorer


Theorem possumd

Description: Condition for a positive sum. (Contributed by Scott Fenton, 16-Dec-2017)

Ref Expression
Hypotheses possumd.1 ( 𝜑𝐴 ∈ ℝ )
possumd.2 ( 𝜑𝐵 ∈ ℝ )
Assertion possumd ( 𝜑 → ( 0 < ( 𝐴 + 𝐵 ) ↔ - 𝐵 < 𝐴 ) )

Proof

Step Hyp Ref Expression
1 possumd.1 ( 𝜑𝐴 ∈ ℝ )
2 possumd.2 ( 𝜑𝐵 ∈ ℝ )
3 2 renegcld ( 𝜑 → - 𝐵 ∈ ℝ )
4 3 1 posdifd ( 𝜑 → ( - 𝐵 < 𝐴 ↔ 0 < ( 𝐴 − - 𝐵 ) ) )
5 1 recnd ( 𝜑𝐴 ∈ ℂ )
6 2 recnd ( 𝜑𝐵 ∈ ℂ )
7 5 6 subnegd ( 𝜑 → ( 𝐴 − - 𝐵 ) = ( 𝐴 + 𝐵 ) )
8 7 breq2d ( 𝜑 → ( 0 < ( 𝐴 − - 𝐵 ) ↔ 0 < ( 𝐴 + 𝐵 ) ) )
9 4 8 bitr2d ( 𝜑 → ( 0 < ( 𝐴 + 𝐵 ) ↔ - 𝐵 < 𝐴 ) )