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Proof step 9 from proof nat (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 10:
0 < j, 0 ≤ i ⊦ i→n(i) / i→n(j) = i→n(i / j)

Rule argument

spec: <toplevel> name: natpos seq: 0 < j, 0 ≤ i ⊦ i→n(i) / i→n(j) = i→n(i / j) subst: [j, i] → [~ j, i]

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i);%j ≠ 0 ⊦ i / ~ j = ~(i / j); (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 < ~ i ↔ i < 0; (s) Used terms: ~ j, i