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Proof step 11 from proof add-scale-neq (
previous step
)
⊦ 0 ≤ j ∧ 0 ≤ j0 ∧ i % 10 ≠ 0 ∧ i0 % 10 ≠ 0 ∧ (i + i0) % 10 ≠ 0 → maxScale(i * 10 ^ j + i0 * 10 ^ j0) = min(j, j0)
A spec-lemma