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Proof step 3 from proof bmin-bneq (previous step)

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

User applied rule insert lemma

And obtained the following new goals

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

Rule argument

spec: <toplevel> name: Min-in seq: ⊦ s ≠ ∅ → s.min ∈ s subst: [s] → [s -- s.min]

Additional information

Used terms: s -- s.min