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Proof step 2 from proof min-sub-01 (previous step)

s0.min ∈ s, s ⊆ s0 ⊦ s.min = s0.min ↔ true, s = ∅

Heuristic simplifier applied rule simplifier

And closed the goal!

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;%min-less: ⊦ a ≠ s.min ∧ a ∈ s → s.min < a;%minmax-04: ⊦ s ≠ ∅ → ¬ s.max < s.min;% ⊦ ¬ a < a; (s)% ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss)%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%sub-in: ⊦ a ∈ s ∧ s ⊆ s0 → a ∈ s0; (forward) from specification set-basic