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Proof step 19 from proof sub-add (previous step)

0 ≤ abs(i0) ∧ 0 ≤ abs(i2) ⊦ (i * 10 ^ abs(i2)) * 10 ^ abs(i2) + (i1 * 10 ^ abs(i0)) * 10 ^ abs(i2) = i * 10 ^ (2 * abs(i2)) + i1 * 10 ^ abs(i0) * 10 ^ abs(i2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
⊦ i = 0, 10 ^ abs(i2) * 10 ^ abs(i2) = 10 ^ (2 * abs(i2))

Rule argument

Additional information

Used simplifier rules: eq-nonzero: ⊦ i ≠ 0 → (i * j = i * j0 ↔ j = j0); 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%add-in: ⊦ i + j = i + i0 ↔ 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