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Proof step 1 from proof min-ins-01 (conclusion)

⊦ a ∈ s ∧ ¬ b < a → (s ++ b).min = s.min

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ s, s.min ∈ s, s.max ∈ s ⊦ s = ∅, (s ++ b).min = s.min, # s = 0, a < s.min, s.max < a, b < a, b < s.min

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;%minmax-04: ⊦ s ≠ ∅ → ¬ s.max < s.min;%irreflexivity: ⊦ ¬ a < a; from specification oelem%nnn: ⊦ ¬ c < b ∧ ¬ a < c → ¬ a < b; (forward) from specification oelem%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic