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Proof step 16 from proof square-ls (previous step)

0 ≤ abs(i2) ⊦ i * i * 10 ^ abs(i2) * 10 ^ abs(i2) < i1 * i1 * 10 ^ (2 * abs(i0)) ↔ abs(mkDecimal(i, i0)) < abs(mkDecimal(i1, i2))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
⊦ i * i * 10 ^ abs(i2) * 10 ^ abs(i2) < i1 * i1 * 10 ^ (2 * abs(i0)) ↔ abs(mkDecimal(i, i0)) < abs(mkDecimal(i1, i2))

Rule argument

Additional information

Used simplifier rules: abs-03: ⊦ 0 ≤ abs(i); from specification int-abs