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Proof step 4 from proof short-in (previous step)

¬(0 ≤ i ∧ i % 65536 ≤ 32767), i ≤ 32767, 0 ≤ i ⊦ s2i(i2s(i)) = i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < i % 65536, i < i % 65536, 32767 < i % 65536, 0 ≤ i, i ≤ 32767 ⊦ i % 65536 = i + 65536

Rule argument

Additional information

Used simplifier rules: sis-pos-high: ⊦ 0 ≤ i ∧ 32767 < i % 65536 → s2i(i2s(i)) = i % 65536 - 65536;%i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;