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Proof step 21 from proof mod-ls-same-01 (previous step)

0 < i ⊦ abs(j) = 1, j = 1, j = 0, i = 0, i % j < i ↔ ¬ i < abs(j)

User applied rule insert elim lemma

And obtained the following new goals

Proof step 22:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0)
Proof step 23:
0 < i ⊦ 0 < j ∧ 0 ≤ i, abs(j) = 1, j = 1, j = 0, i = 0, i % j < i ↔ ¬ i < abs(j)

Rule argument

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

Additional information

Used simplifier rules: abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%ls-add: ⊦ 0 ≤ j1 ∧ j1 < j → (i0 * j + j1 < j ↔ i0 ≤ 0); from specification int-mult%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs Used terms: i, j, i0, j0