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Proof step 8 from proof labsp (previous step)

0 ≤ i0 -1, 0 ≤ j, ∀ j. 0 ≤ j ∧ 0 ≤ i0 → (i→n(i0) < i→n(j) ↔ abs(i0) < abs(j)) ⊦ i→n(i0 -1) < i→n(j) ↔ abs(i0 -1) < abs(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: sub: j ≤ i, 0 ≤ j ⊦ i→n(i - j) = i→n(i) - i→n(j);%abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ i < j + 1 ↔ i ≤ j; (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ 0 < j → abs(j) = j; (s)% ⊦ 0 < i → abs(i - 1) = abs(i) - 1; (s)% ⊦ 0 ≤ i → abs(i) = i; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ i ≤ j → ¬ j < i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ i -1 = i - 1; (s)