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Proof step 1 from proof absls-04 (conclusion)

⊦ i < abs(i -1) ↔ i ≤ 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: sub-pred: ⊦ i -1 = i - 1;%le-def: ⊦ i ≤ j ↔ ¬ j < i;%abs-09: ⊦ 0 < i → abs(i - 1) = abs(i) - 1;%abs-16: ⊦ 0 < j → abs(j) = j;%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j;%abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0;%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j;%sub-eq-01: ⊦ i = j → i - j = 0;%lst-02: ⊦ ¬ i < j → (¬ j < i ↔ i = j); from specification int-basic1%lf-08: ⊦ ¬ i < j ∧ ¬ i0 < i → ¬ i0 < j; (forward) from specification int-basic1%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2