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Proof step 6 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, (s -- s.min).min ≠ s.min ∧ (s -- s.min).min ∈ s

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ s ≠ ∅ ∧ s ⊆ s0 → s.min ∈ s0
Proof step 8:
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, s -- s.min ≠ ∅ ∧ s -- s.min ⊆ s
Proof step 14:
s.max ∈ s, s.min ∈ s, s -- s.min ⊆ s ⊦ # s = 1, # s = 0, s -- s.min = ∅, s = ∅, s.max < (s -- s.min).min, s.min < (s -- s.min).min, s.max < s.min, (s -- s.min).min ≠ s.min ∧ true

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 terms: s, s -- s.min