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Proof step 20 from proof i2np (previous step)

i→n(j) * i→n(i1) + i→n(j) = i→n(i1 * j + 1 * j), 0 ≤ i1, 0 ≤ j, ∀ j0. 0 ≤ j0 → i→n(j0) * i→n(i1) + i→n(j0) = i→n((i1 + 1) * j0) ⊦ i→n(i1) * i→n(j) = i→n(i1 * j)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: add-01 seq: 0 ≤ i, 0 ≤ j ⊦ i→n(i + j) = i→n(i) + i→n(j) subst: [j, i] → [1 * j, i1 * j] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 1 * i = i; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ i * j + i = i * (j + 1); (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ 0 ≤ i → (0 ≤ i * j ↔ i = 0 ∨ 0 ≤ j); (s)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s) Used terms: 1 * j, i1 * j