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Proof step 13 from proof short-larger (previous step)

0 ≤ i ∧ 32767 < i % 65536, 32768 ≤ i, 0 < i, 0 ≤ i ⊦ i % 65536 - 65536 = s2i(i2s(i - 65536))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
0 < i % 65536, 0 ≤ i, 0 < i, 32768 ≤ i, 32767 < i % 65536 ⊦ i % 65536 = s2i(i2s(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)% ⊦ i - j = i0 ↔ i = i0 + j;