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Proof step 8 from proof fillsgn-len-short (previous step)

s2i(sho) < 0, 2 < # int2bytes(s2i(sho)) ⊦ # 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

Proof step 9:
2 < # int2bytes(s2i(sho)), s2i(sho) < 0 ⊦ # fillsgn(2, int2bytes(s2i(sho))) = 2

Rule argument

Additional information

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