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Proof step 14 from proof fillsgn-len-short (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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