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Proof step 3 from proof bmin-del-04 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ s -- a ≠ ∅ ∧ a ≠ s.min → (s -- a).butmin = s.butmin -- a

Rule argument

spec: <toplevel> name: bmin-del-03 seq: ⊦ s -- a ≠ ∅ ∧ a ≠ s.min → (s -- a).butmin = s.butmin -- a subst: [a, s] → [a, s] flags: <all paths>

Additional information

Used simplifier rules: minmax-04: ⊦ s ≠ ∅ → ¬ s.max < s.min;%Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min;% ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; (ws)% ⊦ s = ∅ ++ a → s -- a = ∅; (s)%one-size: ⊦ s = ∅ ++ a → # s = 1; (forward) from specification set-basic%in-06: ⊦ a ∈ s ∧ s = ∅ ++ b → a = b; (forward) from specification set-basic Used terms: a, s