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Proof step 18 from proof short-larger (previous step)

0 ≤ i - 65536 ∧ 32767 < (i - 65536) % 65536, 0 < i % 65536, 0 ≤ i, 0 < i, 32768 ≤ i, 32767 < i % 65536 ⊦ i % 65536 = ((i - 65536) % 65536 - 65536) + 65536

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
0 < (i - 65536) % 65536, 65536 ≤ i, 32767 < i % 65536, 0 < i, 0 ≤ i, 0 < i % 65536, 32767 < (i - 65536) % 65536 ⊦ i % 65536 = (i - 65536) % 65536

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)%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (j - i) = j;