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Proof step 8 from proof pot-zero-01 (previous step)

j < i, j < 0, 0 < i, j + 1 < i ⊦ 1 / i ^ ~ j = 0 ↔ 1 < i

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: div-zero seq: 0 < j, 0 ≤ i, i < j ⊦ i / j = 0 subst: [j, i] → [i ^ ~ j, 1] flags: rewrite

Additional information

Used simplifier rules: grz: ⊦ 0 < i → (0 < i ^ j ↔ i = 1 ∨ 0 ≤ j);%eqone-02: ⊦ i ^ ~ j = 1 ∧ j < 0 → abs(i) = 1;%one-01: ⊦ 1 ^ j = 1;%numlelsdel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%negle-07: ⊦ i ≤ ~ j ↔ i + j ≤ 0; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%div-eq-zero: ⊦ 0 < i ∧ 0 < j → (i / j = 0 ↔ i < j); from specification int-div%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs Used terms: i ^ ~ j, 1