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Proof step 6 from proof fillsgn-len-short (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
2 < # int2bytes(s2i(sho)) ⊦ # fillsgn(2, int2bytes(s2i(sho))) = 2, ¬ s2i(sho) < 0 ∧ int2bytes(s2i(sho)).first ≠ i2b(0)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;