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Proof step 3 from proof min-delmin (previous step)

s.min ∈ s, s.max ∈ s ⊦ # s = 0, # s = 1, s = ∅, s.max < s.min, s.min < (s -- s.min).min, s.max < (s -- s.min).min

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ a ≠ s.min ∧ a ∈ s → s.min < a
Proof step 5:
s.max ∈ s, s.min ∈ s ⊦ # s = 1, # s = 0, s = ∅, s.max < (s -- s.min).min, s.min < (s -- s.min).min, s.max < s.min, (s -- s.min).min ≠ s.min ∧ (s -- s.min).min ∈ s

Rule argument

spec: <toplevel> name: min-less seq: ⊦ a ≠ s.min ∧ a ∈ s → s.min < a subst: [a, s] → [(s -- s.min).min, s] flags: <all paths>

Additional information

Used terms: (s -- s.min).min, s