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Proof step 35 from proof ex-modeq-half (previous step)

j ≠ 0, ∃ j0. i0 % j - j = j0 * j, 0 ≤ i0 % j, j / 2 < i0 % j, 0 < j, odd(j) ⊦ (i0 % j - j) |%| j = i0 % j ∧ j ≤ i0 % j + j / 2, ∃ i. ≡mod(i, i0, j) ∧ abs(i) ≤ j / 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
i0 % j = j0 * j + j, 0 ≤ i0 % j, j / 2 < i0 % j, 0 < j, odd(j) ⊦ j = 0, (i0 % j - j) |%| j = i0 % j ∧ j ≤ i0 % j + j / 2, ∃ i. ≡mod(i, i0, j) ∧ abs(i) ≤ j / 2

Rule argument

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs