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Proof step 2 from proof short-add-max-01 (previous step)

s2i(sho1) < 0, 0 < s2i(sho0) + s2i(sho1), 32767 < s2i(sho0) + s2i(sho1) ⊦

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: 32767 < s2i(sho0) + 0

Additional information

Used simplifier rules: short-ngr: ⊦ ¬ 32767 < s2i(sho);% ⊦ i + 0 = i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used formulas: 32767 < s2i(sho0) + 0