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Proof step 5 from proof bmin-two (previous step)

∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b ⊦ s.butmin.min ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ (∅ ++ a ++ b).butmin.min = b, (∅ ++ a ++ b).butmin.min = a, a = b

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ ∅ ++ b ↔ a = b; (s)% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (ws)