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Proof step 22 from proof square-isSquare (previous step)

¬ 2 * abs(i0) - 1 < 0, ∀ i1, i0. i→n(i1) < i→n(j * 10) → true → isSquare(mkDecimal(i1 * i1, 2 * abs(i0)).int) ⊦ i0 = 0, j = 0, isSquare(mkDecimal(j * j, 2 * abs(i0) - 1 - 1).int)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%abs-zero: ⊦ abs(i) ≤ 0 ↔ i = 0; from specification int-abs%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%nummult-ls: ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i); from specification int-div%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; 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%sub-sub: ⊦ i - j - i0 = i - j + i0; from specification int-abs