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Proof step 69 from proof sub-def-any (previous step)

mkDecimal(j1, abs(j) - 1) - mkDecimal(i0, j0) = mkDecimal((j1 * 10 ^ abs(j0) - i0 * 10 ^ (abs(j) - 1)) * 10, 2 - abs(j0) + abs(j)), abs(j0) + abs(j) ≤ 0, ∀ 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, mkDecimal((j1 * 10 ^ abs(j0) - i0 * 10 ^ (abs(j) - 1)) * 10, 2 - abs(j0) + abs(j)) = mkDecimal((j1 * 10) * 10 ^ abs(j0) - i0 * 10 ^ abs(j), abs(j) + abs(j0))

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: lels-02 seq: ⊦ i ≠ j → (i ≤ j ↔ i < j) subst: [j, i] → [0, abs(j0) + abs(j)] flags: rewrite

Additional information

Used simplifier rules: abs-add-pos: ⊦ ¬ abs(i) + abs(j) < 0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%abs-add-zero: ⊦ abs(i) + abs(j) = 0 ↔ i = 0 ∧ j = 0; from specification int-abs Used terms: 0, abs(j0) + abs(j)