Metamath Proof Explorer


Theorem eqleltd

Description: Equality in terms of 'less than or equal to', 'less than'. (Contributed by NM, 7-Apr-2001)

Ref Expression
Hypotheses ltd.1 ( 𝜑𝐴 ∈ ℝ )
ltd.2 ( 𝜑𝐵 ∈ ℝ )
Assertion eqleltd ( 𝜑 → ( 𝐴 = 𝐵 ↔ ( 𝐴𝐵 ∧ ¬ 𝐴 < 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 ltd.1 ( 𝜑𝐴 ∈ ℝ )
2 ltd.2 ( 𝜑𝐵 ∈ ℝ )
3 eqlelt ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 = 𝐵 ↔ ( 𝐴𝐵 ∧ ¬ 𝐴 < 𝐵 ) ) )
4 1 2 3 syl2anc ( 𝜑 → ( 𝐴 = 𝐵 ↔ ( 𝐴𝐵 ∧ ¬ 𝐴 < 𝐵 ) ) )