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Proof step 12 from proof i2np (previous step)

0 ≤ j, 0 ≤ i0, ∀ j0. 0 ≤ j0 → i→n(i0) * i→n(j0) = i→n(i0 * j0) ⊦ i→n(j) * i→n(i0) = i→n(i0 * j)

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: j abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: j