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Proof step 4 from proof bmin-two-07 (
previous step
)
s.butmin = ∅, ∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b ⊦
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 5
:
(∅ ++ a ++ b).butmin = ∅ ⊦ a = b
Rule argument
Additional information