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Proof step 7 from proof fillsgn-len-short (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 8:
s2i(sho) < 0, 2 < # int2bytes(s2i(sho)) ⊦ # fillsgn(2, int2bytes(s2i(sho))) = 2, ¬ s2i(sho) < 0 ∧ int2bytes(s2i(sho)).first ≠ i2b(0)
Proof step 14:
2 < # int2bytes(s2i(sho)) ⊦ s2i(sho) < 0, # fillsgn(2, int2bytes(s2i(sho))) = 2, ¬ s2i(sho) < 0 ∧ int2bytes(s2i(sho)).first ≠ i2b(0)

Rule argument

fma: s2i(sho) < 0

Additional information

Used formulas: s2i(sho) < 0