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Proof step 1 from proof short-size-02 (conclusion)

⊦ ¬ 16 < # int2bits(s2i(sho))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
16 < # int2bits(s2i(sho)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)