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Proof step 3 from proof bmin-sub-01 (previous step)

# s = 2, s ⊆ s0 ⊦ s.butmin.min ∈ s0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ s ≠ ∅ ∧ s ⊆ s0 → s.min ∈ s0
Proof step 5:
# s = 2, s ⊆ s0 ⊦ s.butmin.min ∈ s0, s.butmin ≠ ∅ ∧ s.butmin ⊆ s0

Rule argument

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

Additional information

Used terms: s0, s.butmin