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Proof step 4 from proof mod2-add-neg (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j)
Proof step 6:
j = i0 * 2 + j0 ∧ 0 ≤ j0 ∧ j0 < 2 ∧ 0 < 2 ∧ 0 ≤ j, j0 = 0, abs(i) ≤ j, i ≤ 0, 0 < j, i < j ⊦ i % 2 + (i + j) % 2 = 0

Rule argument

spec: int-div inst: 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: rewrite, current

Additional information

Used simplifier rules: lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: j, 2, i0, j0