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Proof step 7 from proof labsp (previous step)

∀ j. 0 ≤ i0 ∧ 0 ≤ j → (i→n(i0) < i→n(j) ↔ abs(i0) < abs(j)), 0 ≤ i0 -1, 0 ≤ 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%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%abs-09: ⊦ 0 < i → abs(i - 1) = abs(i) - 1; from specification int-abs%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ 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%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%sub-pred: ⊦ i -1 = i - 1; from specification int-abs