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Proof step 4 from proof natpos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i→n(i0 * j + j0) = m0 * i→n(j) + n0, 0 < j, j0 < j, n0 < i→n(j), 0 ≤ j0, 0 ≤ i0 ⊦ j = 0, m0 = i→n(i0)

Rule argument

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic