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Proof step 5 from proof one-neg (previous step)

1 / j < 0, j < 0 ⊦ j = 0, j = ~1

User applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
⊦ j < 0 ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < abs(j))
Proof step 7:
1 = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < abs(j) ∧ 0 ≤ 1, i0 < 0, j < 0 ⊦ j = 0, j = ~1

Rule argument

spec: int-div inst: name: div-elim1 seq: ⊦ j < 0 ∧ 0 ≤ i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < abs(j)) subst: [i, j, i0, j0] → [1, j, i, j0] flags: rewrite, current

Additional information

Used terms: 1, j, i, j0