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Proof step 5 from proof ex-modeq-half (previous step)

0 < j, odd(j), 0 < j ⊦ (i0 |%| j) |%| j = i0 |%| j ∧ abs(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 6:
odd(j), 0 < j, j / 2 < abs(i0 |%| j) ⊦ ∃ i. ≡mod(i, i0, j) ∧ abs(i) ≤ j / 2

Rule argument

Additional information

Used simplifier rules: mod-mod: ⊦ 0 < j → (i |%| j) |%| j = i |%| j;%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs