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Proof step 11 from proof div-add-nosucc (previous step)

(j0 * (i1 + 1) + i1) / (i1 + 1) = i1 / (i1 + 1) + j0, 0 ≤ j0, 0 < i1 ⊦ ~1 = i1, i1 / (i1 + 1) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: div-eq-zero seq: ⊦ 0 < i ∧ 0 < j → (i / j = 0 ↔ i < j) subst: [j, i] → [i1 + 1, i1] flags: rewrite, current

Additional information

Used simplifier rules: zero-02: ⊦ 0 < i ∧ 0 < j → 0 < i + j; 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%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%irref: ⊦ ¬ i < i; from specification int-basic1 Used terms: i1 + 1, i1