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Proof step 49 from proof add-def-any (previous step)

(10 * (j1 * 10 ^ abs(j0) + i0 * 10 ^ (abs(j) - 1))) % 10 = 0 ∧ abs(j) + abs(j0) ≠ 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, abs(j) - 1) + mkDecimal(i0, j0) = mkDecimal((10 * (j1 * 10 ^ abs(j0) + i0 * 10 ^ (abs(j) - 1))) / 10, abs(abs(j) + abs(j0)) - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%abs-abs-01: ⊦ abs(abs(i) + abs(j)) = abs(i) + abs(j); from specification int-abs%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%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%mod-zero-num: i0 % j = 0, j ≠ 0 ⊦ (i * (* i0)) % (* j) = 0; from specification int-div