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Proof step 4 from proof short-2bytes-01 (previous step)

¬ bytes2int(by ' + by0 ') ≤ 32767, 65536 ≤ bytes2int(by ' + by0 ') + s2i(sho), 0 < bytes2int(by ' + by0 ') + s2i(sho), 0 ≤ bytes2int(by ' + by0 ') + s2i(sho) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < bytes2int(by ' + by0 '), 32767 < bytes2int(by ' + by0 '), 0 ≤ bytes2int(by ' + by0 ') + s2i(sho), 0 < bytes2int(by ' + by0 ') + s2i(sho), 65536 ≤ bytes2int(by ' + by0 ') + s2i(sho) ⊦

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ ¬ i ≤ j ↔ j < i;