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Proof step 14 from proof short-size (previous step)

~1 = s2i(sho), 16 < # int2bits(s2i(sho)), s2i(sho) ≤ 0, ~1 ≤ s2i(sho) ⊦ # int2bits(s2i(sho)) - 1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
~1 = s2i(sho), 16 < # int2bits(~1) ⊦ # int2bits(~1) - 1 = 0

Rule argument

Additional information

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