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

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, j0 < j, n0 < i→n(j), i→n(j) ≤ i→n(j0), n0 < i→n(j0), 0 ≤ i0, 1 < j, 0 < j, 0 < j0 ⊦ j = 0, m0 = i→n(i0), j0 = 0

Rule argument

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat