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Proof step 11 from proof fillsgn-len-short (previous step)

¬(s2i(sho) < 0 ∧ int2bytes(s2i(sho)).first ≠ i2b(~1)), 2 < # int2bytes(s2i(sho)), s2i(sho) < 0 ⊦ # fillsgn(2, int2bytes(s2i(sho))) = 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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