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Proof step 3 from proof nat (previous step)

j < 0, i < 0, j ≠ 0 ⊦ i→n(i) / i→n(j) = i→n(i / j)

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: intnat inst: name: natpos seq: 0 < j, 0 ≤ i ⊦ i→n(i) / i→n(j) = i→n(i / j) subst: [j, i] → [~ j, ~ i] flags: current, []

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i);%neg-10: ⊦ ~ i = 0 ↔ i = 0; from specification int-abs%neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j); from specification int-div%neg: j ≠ 0 ⊦ i / ~ j = ~(i / j); from specification int-div%neg-02: ⊦ ~ ~ i = i; from specification int-abs%neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs%zero-gels-01: ⊦ 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%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs Used terms: ~ j, ~ i