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Proof step 25 from proof mult-def-any (previous step)

10 ≠ 0, ∃ j0. i0 = j0 * 10, ∀ i1, i2, j, j0. i→n(i1) + i→n(i2) < i→n(i) + i→n(i0) → true → mkDecimal(i1, j) * mkDecimal(i2, j0) = mkDecimal(i1 * i2, abs(j) + abs(j0)) ⊦ j0 = 0, mkDecimal(i, j) * mkDecimal(i0 / 10, abs(j0) - 1) = mkDecimal(i * i0, abs(j) + abs(j0)), i % 10 = 0 ∧ j ≠ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
∀ 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)) ⊦ j0 = 0, mkDecimal(i, j) * mkDecimal(j1, abs(j0) - 1) = mkDecimal(i * j1 * 10, abs(j) + abs(j0)), i % 10 = 0 ∧ j ≠ 0

Rule argument

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%div-mult-num: j ≠ 0, j0 % j = 0 ⊦ (i * (* j0)) / (* j) = i * j0 / j; from specification int-div