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Proof step 13 from proof ushort-pos-02 (previous step)

i < 0 ∧ ~32768 ≤ i % 65536, 0 ≤ i % 65536, i < 0, ~65536 ≤ i, i < ~32768 ⊦ i % 65536 = i + 65536

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
i % 65536 = 0, i < 0, i < ~32768, ~65536 < i ⊦ ~65536 = i

Rule argument

Additional information

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