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Proof step 3 from proof min-delmin-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ s ≠ ∅ ∧ s ⊆ s0 → s.min ∈ s0

Rule argument

spec: <toplevel> name: min-subset seq: ⊦ s ≠ ∅ ∧ s ⊆ s0 → s.min ∈ s0 subst: [s0, s] → [s, s -- s.min] flags: <all paths>

Additional information

Used simplifier rules: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; (ws)% ⊦ s -- a ⊆ s; (s) Used terms: s, s -- s.min