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Proof step 4 from proof natpos (previous step)

i→n(i0 * j + j0) = m0 * i→n(j) + n0 ∧ n0 < i→n(j) ∧ i→n(j) ≠ 0, j0 < j, 0 ≤ j0, 0 < j, 0 ≤ i0 ⊦ 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 ≤ i0, 0 < j, 0 ≤ j0, j0 < j, n0 < i→n(j) ⊦ 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-basic1