Metamath Proof Explorer


Theorem resubval

Description: Value of real subtraction, which is the (unique) real x such that B + x = A . (Contributed by Steven Nguyen, 7-Jan-2023)

Ref Expression
Assertion resubval ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 𝐵 ) = ( 𝑥 ∈ ℝ ( 𝐵 + 𝑥 ) = 𝐴 ) )

Proof

Step Hyp Ref Expression
1 eqeq2 ( 𝑦 = 𝐴 → ( ( 𝑧 + 𝑥 ) = 𝑦 ↔ ( 𝑧 + 𝑥 ) = 𝐴 ) )
2 1 riotabidv ( 𝑦 = 𝐴 → ( 𝑥 ∈ ℝ ( 𝑧 + 𝑥 ) = 𝑦 ) = ( 𝑥 ∈ ℝ ( 𝑧 + 𝑥 ) = 𝐴 ) )
3 oveq1 ( 𝑧 = 𝐵 → ( 𝑧 + 𝑥 ) = ( 𝐵 + 𝑥 ) )
4 3 eqeq1d ( 𝑧 = 𝐵 → ( ( 𝑧 + 𝑥 ) = 𝐴 ↔ ( 𝐵 + 𝑥 ) = 𝐴 ) )
5 4 riotabidv ( 𝑧 = 𝐵 → ( 𝑥 ∈ ℝ ( 𝑧 + 𝑥 ) = 𝐴 ) = ( 𝑥 ∈ ℝ ( 𝐵 + 𝑥 ) = 𝐴 ) )
6 df-resub = ( 𝑦 ∈ ℝ , 𝑧 ∈ ℝ ↦ ( 𝑥 ∈ ℝ ( 𝑧 + 𝑥 ) = 𝑦 ) )
7 riotaex ( 𝑥 ∈ ℝ ( 𝐵 + 𝑥 ) = 𝐴 ) ∈ V
8 2 5 6 7 ovmpo ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 𝐵 ) = ( 𝑥 ∈ ℝ ( 𝐵 + 𝑥 ) = 𝐴 ) )