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Proof step 39 from proof sub-def-any (previous step)

¬ abs(j) + abs(j0) - 1 < 0, mkDecimal(i, j) - mkDecimal(j1, abs(j0) - 1) = mkDecimal((i * 10 ^ (abs(j0) - 1) - j1 * 10 ^ abs(j)) * 10, (abs(j) + abs(j0) - 1) + 1), ∀ 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 * 10 ^ abs(j0) - i2 * 10 ^ abs(j), abs(j) + abs(j0)) ⊦ j0 = 0, mkDecimal((i * 10 ^ (abs(j0) - 1) - j1 * 10 ^ abs(j)) * 10, (abs(j) + abs(j0) - 1) + 1) = mkDecimal(i * 10 ^ abs(j0) - (j1 * 10) * 10 ^ abs(j), abs(j) + abs(j0)), i % 10 = 0 ∧ j ≠ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 40:
mkDecimal(i, j) - mkDecimal(j1, abs(j0) - 1) = mkDecimal((i * 10 ^ (abs(j0) - 1) - j1 * 10 ^ abs(j)) * 10, abs(j) + abs(j0)), 0 < abs(j) + abs(j0), ∀ 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 * 10 ^ abs(j0) - i2 * 10 ^ abs(j), abs(j) + abs(j0)) ⊦ j0 = 0, i * 10 ^ abs(j0) = (i * 10 ^ (abs(j0) - 1) - j1 * 10 ^ abs(j)) * 10 + (j1 * 10) * 10 ^ abs(j), i % 10 = 0 ∧ j ≠ 0

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%eq-same: ⊦ mkDecimal(i, j) = mkDecimal(i0, j) ↔ i = i0; from specification decimal%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%Sub-def: ⊦ i + (j - i) = j; from specification int-abs