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Proof step 5 from proof bmin-sub-01 (previous step)

# s = 2, s ⊆ s0 ⊦ s.butmin.min ∈ s0, s.butmin ≠ ∅ ∧ s.butmin ⊆ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# s = 2, s ⊆ s0 ⊦ s.butmin.min ∈ s0

Rule argument

Additional information

Used simplifier rules: min-sub: ⊦ ¬ s.min ∈ s0 → (s ⊆ s0 ↔ s = ∅);