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Proof step 1 from proof min-two-02 (conclusion)

⊦ # s = 2 ∧ a ∈ s ∧ a ≠ s.min → ∅ ++ a ++ s.min = s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%min-less: ⊦ a ≠ s.min ∧ a ∈ s → s.min < a;%Max-in: ⊦ s ≠ ∅ → s.max ∈ s;%Min-in: ⊦ s ≠ ∅ → s.min ∈ s;%minmax-04: ⊦ s ≠ ∅ → ¬ s.max < s.min;%irreflexivity: ⊦ ¬ a < a; from specification oelem%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%sls: ⊦ a ≠ b → (¬ a < b ↔ b < a); from specification oelem%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic