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Proof step 1 from proof ex-modeq-half (conclusion)

0 < j ∧ odd(j) ⊦ ∃ i. ≡mod(i, i0, j) ∧ abs(i) ≤ j / 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
odd(j), 0 < j ⊦ ∃ i. ≡mod(i, i0, j) ∧ abs(i) ≤ j / 2

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs