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Proof step 24 from proof mult-dist (previous step)

0 ≤ abs(i0) ∧ 0 ≤ abs(i2) ⊦ (i * (i1 * 10 ^ abs(i4) + i3 * 10 ^ abs(i2))) * 10 ^ abs(i0) = i * i1 * 10 ^ (abs(i0) + abs(i4)) + i * i3 * 10 ^ abs(i0) * 10 ^ abs(i2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
⊦ i1 = 0, i = 0, 10 ^ abs(i4) * 10 ^ abs(i0) = 10 ^ (abs(i0) + abs(i4))

Rule argument

Additional information

Used simplifier rules: eq-nonzero: ⊦ i ≠ 0 → (i * j = i * j0 ↔ j = j0); from specification int-mult%zero-eq: ⊦ i * j = 0 ↔ i = 0 ∨ j = 0; from specification int-mult%neq-not-zero: ⊦ i * j ≠ i * j0 → i ≠ 0; (forward) from specification int-mult%abs-03: ⊦ 0 ≤ abs(i); from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%eq-dist: ⊦ (i + j) * i0 = j0 + j * i0 ↔ i * i0 = j0; from specification int-mult