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Proof step 34 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)) ⊦ abs(j) + abs(j0) = 0, j0 = 0, mkDecimal(i, j) + mkDecimal(j1, abs(j0) - 1) = mkDecimal(i * 10 ^ (abs(j0) - 1) + j1 * 10 ^ abs(j), abs(j) + abs(j0) - 1), i % 10 = 0 ∧ j ≠ 0

User applied rule apply induction

And obtained the following new goals

Proof step 35:
∀ 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)) ⊦ i→n(i) + i→n(j1) < i→n(i) + i→n(j1 * 10), abs(j) + abs(j0) = 0, j0 = 0, mkDecimal(i, j) + mkDecimal(j1, abs(j0) - 1) = mkDecimal(i * 10 ^ (abs(j0) - 1) + j1 * 10 ^ abs(j), abs(j) + abs(j0) - 1), i % 10 = 0 ∧ j ≠ 0

Rule argument

subst: [i1, i2, j, j0] → [i, j1, j, abs(j0) - 1]

Additional information

Used simplifier rules: abs-abs: ⊦ j ≠ 0 → abs(abs(j) - 1) = abs(j) - 1; from specification int-abs%sub-add-01: ⊦ i + (j - i0) = i + j - i0; 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 Used terms: i, j1, j, abs(j0) - 1