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Proof step 2 from proof min-two-02 (previous step)

# s = 2, a ∈ s ⊦ ∅ ++ a ++ s.min = s, a = s.min

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# s = 2, s.min < a, a ∈ s, s.min ∈ s, s.max ∈ s ⊦ ∅ ++ a ++ s.min = s, s = ∅, s.max < s.min, s.max < a

Rule argument

Additional information

Used simplifier rules: Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%Max-in: ⊦ s ≠ ∅ → s.max ∈ s;%Min-in: ⊦ s ≠ ∅ → s.min ∈ s;% ⊦ ¬ a < a; (s)%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic% ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic