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Proof step 15 from proof ushort-short (previous step)

¬ abs(s2i(sho) + 65536) < 65536, 0 < (s2i(sho) + 65536) % 65536, s2i(sho) < 0, 32767 < (s2i(sho) + 65536) % 65536, ~65536 ≤ s2i(sho) ⊦ (s2i(sho) + 65536) % 65536 = s2i(sho) + 65536

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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 < abs(i) ↔ i ≠ 0;% ⊦ 0 ≤ abs(i);% ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;