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

(∅ ++ a).butmin.min ∈ (∅ ++ a).butmin, (∅ ++ a).butmin.max ∈ (∅ ++ a).butmin ⊦ # (∅ ++ a).butmin = 0, (∅ ++ a).butmin = ∅, (∅ ++ a).butmin.max < (∅ ++ a).butmin.min

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ s ≠ ∅ → s -- s.min = s.butmin

Rule argument

spec: <toplevel> name: bmin-del-05 seq: ⊦ s ≠ ∅ → s -- s.min = s.butmin subst: [s] → [∅ ++ a] flags: <rotated> <all paths>

Additional information

Used simplifier rules: min-one: ⊦ (∅ ++ a).min = a;% ⊦ (s ++ a) -- a = s -- a; (s)% ⊦ ∅ -- a = ∅; (s)% ⊦ ∅ ≠ s ++ a; (s) Used terms: ∅ ++ a