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Proof step 29 from proof add-def-any (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 30:
i * 10 ^ (abs(j0) - 1) * 10 + (j1 * 10) * 10 ^ abs(j) = 10 * (i * 10 ^ (abs(j0) - 1) + j1 * 10 ^ abs(j)), ∀ 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, j) + mkDecimal(j1, abs(j0) - 1) = mkDecimal(i * 10 ^ (abs(j0) - 1) * 10 + (j1 * 10) * 10 ^ abs(j), abs(j) + abs(j0)), i % 10 = 0 ∧ j ≠ 0

Rule argument

fma: i * 10 ^ (abs(j0) - 1) * 10 + (j1 * 10) * 10 ^ abs(j) = 10 * (i * 10 ^ (abs(j0) - 1) + j1 * 10 ^ abs(j))

Additional information

Used simplifier rules: eq-dist: ⊦ (i + j) * i0 = j0 + j * i0 ↔ i * i0 = j0; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: i * 10 ^ (abs(j0) - 1) * 10 + (j1 * 10) * 10 ^ abs(j) = 10 * (i * 10 ^ (abs(j0) - 1) + j1 * 10 ^ abs(j))