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Proof step 2 from proof i2n-mult (previous step)

0 ≤ i, 0 ≤ j ⊦ i→n(i * j) = i→n(i) * i→n(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
0 ≤ i, 0 ≤ j ⊦ i→n(i) * i→n(j) = i→n(i * j)

Rule argument

spec: intnat inst: name: i2np seq: 0 ≤ i, 0 ≤ j ⊦ i→n(i) * i→n(j) = i→n(i * j) subst: [j, i] → [j, i] flags: rewrite, current

Additional information

Used simplifier rules: zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs Used terms: j, i