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Proof step 13 from proof i2n-mod (previous step)

j0 < j, 0 < j, 0 ≤ j0, 0 ≤ i0, 0 ≤ j, i→n(j) ≤ i→n(j0) ⊦ i→n(j0) = i→n(j0) % i→n(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: intnat inst: name: labs-01 seq: ⊦ i→n(i) ≤ i→n(j) ↔ abs(i) ≤ abs(j) subst: [j, i] → [j0, j] flags: rewrite, current

Additional information

Used simplifier rules: abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: j0, j