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Proof step 14 from proof natpos (previous step)

i→n(i0) * i→n(j) + i→n(j0) = m0 * i→n(j) + n0, 1 < j, n0 < i→n(j0), n0 < i→n(j), j0 < j, 0 < j, 0 < j0, i→n(j) ≤ i→n(j0), 0 ≤ i0, 2 ≤ j ⊦ j = 0, j0 = 0, m0 = i→n(i0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ i→n(i) ≤ i→n(j) ↔ abs(i) ≤ abs(j)

Rule argument

spec: <toplevel> name: labs-01 seq: ⊦ i→n(i) ≤ i→n(j) ↔ abs(i) ≤ abs(j) subst: [j, i] → [j0, j] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 < j → abs(j) = j; (s)% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: j0, j