This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.

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 −ℝ 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 ⊢ −ℝ = ( 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 −ℝ 𝐵 ) = ( ℩ 𝑥 ∈ ℝ ( 𝐵 + 𝑥 ) = 𝐴 ) )