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Proof step 13 from proof natpos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
i→n(i0) * i→n(j) + i→n(j0) = m0 * i→n(j) + n0, 1 < j, n0 < i→n(j0), n0 < i→n(j), j0 < j, 0 < j, 0 < j0, i→n(j) ≤ i→n(j0), 0 ≤ i0, 2 ≤ j ⊦ j = 0, j0 = 0, m0 = i→n(i0)

Rule argument

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ 1 ≤ i ↔ 0 < i; (s)%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)