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

s.max ∈ s, s.min ∈ s, s.max ∈ s0, s.min ∈ s0, s0.min ∈ s0, s0.max ∈ s0, s ⊆ s0 ⊦ # s = 0, # s0 = 0, s = ∅ ++ s.min, s = ∅, s0 = ∅, s.min < s0.min, s0.max < s.min, s.max < s0.min, s0.max < s.max, s.max < s.min, s0.max < s0.min, (s -- s.min).min ∈ s0

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] → [s0, s -- s.min] flags: <all paths>

Additional information

Used simplifier rules: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; (ws)% ⊦ s ⊆ s0 → s -- a ⊆ s0; (s) Used terms: s0, s -- s.min