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Proof step 5 from proof mod2-add-neg (previous step)

j % 2 = 0, i < j, 0 < j, abs(i) ≤ j, i ≤ 0 ⊦ i % 2 + (i + j) % 2 = 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j)
Proof step 7:
j = i0 * 2 + j0, j0 = 0, 0 < j, i < j, j0 < 2, 0 < 2, i ≤ 0, abs(i) ≤ j, 0 ≤ j0, 0 ≤ j ⊦ i % 2 + (i + j) % 2 = 0

Rule argument

spec: <toplevel> name: div-elim-pos seq: ⊦ 0 < j ∧ 0 ≤ i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j) subst: [i, j, i0, j0] → [j, 2, i0, j0] flags: <rotated> <all paths>

Additional information

Used simplifier rules: ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s) Used terms: j, 2, i0, j0