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Proof step 9 from proof short-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
16 < # int2bits(s2i(sho)), s2i(sho) ≤ 0 ⊦ # int2bits(s2i(sho)) - 1 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ ¬ i < j ↔ j ≤ i;