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Proof step 80 from proof sub-def-any (previous step)

0 < abs(j0), mkDecimal(j1, abs(j) - 1) - mkDecimal(i0, j0) = mkDecimal((j1 * 10 ^ abs(j0) -1 * 10 - i0 * 10 ^ (abs(j) - 1)) * 10, abs(j0) + abs(j)), 0 < abs(j0) + abs(j), ∀ i1, i2, j, j0. i→n(i1) + i→n(i2) < i→n(j1 * 10) + i→n(i0) → true → mkDecimal(i1, j) - mkDecimal(i2, j0) = mkDecimal(i1 * 10 ^ abs(j0) - i2 * 10 ^ abs(j), abs(j) + abs(j0)) ⊦ j = 0, (j1 * 10) * 10 ^ abs(j0) -1 * 10 = (j1 * 10 ^ abs(j0) -1 * 10 - i0 * 10 ^ (abs(j) - 1)) * 10 + i0 * 10 ^ abs(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 81:
mkDecimal(j1, abs(j) - 1) - mkDecimal(i0, j0) = mkDecimal((j1 * 10 ^ (abs(j0) - 1) * 10 - i0 * 10 ^ (abs(j) - 1)) * 10, abs(j0) + abs(j)), 0 < abs(j0) + abs(j), ∀ i1, i2, j, j0. i→n(i1) + i→n(i2) < i→n(j1 * 10) + i→n(i0) → true → mkDecimal(i1, j) - mkDecimal(i2, j0) = mkDecimal(i1 * 10 ^ abs(j0) - i2 * 10 ^ abs(j), abs(j) + abs(j0)) ⊦ j = 0, 100 * j1 * 10 ^ (abs(j0) - 1) = (j1 * 10 ^ (abs(j0) - 1) * 10 - i0 * 10 ^ (abs(j) - 1)) * 10 + i0 * 10 ^ abs(j), j0 = 0

Rule argument

Additional information

Used simplifier rules: abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%sub-pred: ⊦ i -1 = i - 1; from specification int-abs