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Proof step 2 from proof bmin-in (previous step)

⊦ a ∈ s ↔ a = s.min ∧ s ≠ ∅, a ∈ s.butmin

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ins-inbutmins: ⊦ a ∈ s → (a ∈ s.butmin ↔ a ≠ s.min);%Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%min-inn: ⊦ ¬ s.min ∈ s ↔ s = ∅;%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic