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Proof step 2 from proof sublist-not-empty-01 (previous step)

⊦ sublist(m0, n0, x) = [] ↔ false, n0 = 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));%firstN-empty: ⊦ firstn(n, x) = [] ↔ n = 0;