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Proof step 7 from proof short-nls (previous step)

65536 * i < s2i(sho) ∧ s2i(sho) < 32768 → 65536 * i < 32768, s2i(sho) < 32768, 65536 * i < s2i(sho), 0 < i ⊦

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i);% ⊦ i < 1 ↔ i ≤ 0;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;