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Proof step 30 from proof mult-def-any (previous step)

∀ i1, i2, j, j0. i→n(i1) + i→n(i2) < i→n(i) + i→n(j1 * 10) → true → mkDecimal(i1, j) * mkDecimal(i2, j0) = mkDecimal(i1 * i2, abs(j) + abs(j0)) ⊦ i→n(i) + i→n(j1) < i→n(i) + i→n(j1 * 10), abs(j) + abs(j0) = 0, j0 = 0, mkDecimal(i, j) * mkDecimal(j1, abs(j0) - 1) = mkDecimal(i * j1, abs(j) + abs(j0) - 1), i % 10 = 0 ∧ j ≠ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: Mult-zero: ⊦ i * 0 = 0; from specification int-mult%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%le-same-04: ⊦ i * abs(j) ≤ abs(j) ↔ j = 0 ∨ i ≤ 1; from specification int-mult%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%abs-eqall: ⊦ abs(i * j) = abs(i) * abs(j); from specification int-mult%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2