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Proof step 15 from proof i2np (previous step)

i0 = i1 + 1, 0 < i0, 0 ≤ j, ∀ j. 0 ≤ j ∧ 0 ≤ i0 → i→n(i0) * i→n(j) = i→n(i0 * j) ⊦ (i→n(i0) - 1) * i→n(j) = i→n(i1 * j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: rax: ⊦ 0 < i + 1 → i→n(i + 1) = i→n(i) + 1;%rax-04: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m * (n + 1) = m * n + m; (s)%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; (s)% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ 0 + m = m; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2