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Proof step 1 from proof one (conclusion)

⊦ i / 1 = i

User applied rule insert elim lemma

And obtained the following new goals

Proof step 2:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j)
Proof step 3:
⊦ 0 < 1 ∧ 0 ≤ i, i / 1 = i

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] → [i, 1, i0, j] flags: rewrite, current

Additional information

Used simplifier rules: one: ⊦ 1 * i = i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%eq: ⊦ i = i + j ↔ j = 0; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%num-between: i + 1 = i0 ⊦ j < (* i0) → ¬ (* i) < j; from specification int-basic2 Used terms: i, 1, i0, j