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Proof step 13 from proof ushort-neg (previous step)

¬ # int2bits(s2i(sho)) ≤ 16, s2i(sho) < 0 ⊦ fillsgn(16, int2bits(s2i(sho))) = [], 2 ^ n→i(# fillsgn(16, int2bits(s2i(sho)))) = 65536

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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