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Proof step 4 from proof abs-sub-01 (previous step)

j = i0 + (i + 1) + 1, i < j ⊦ i + 1 = j, abs(i0) < abs(j - i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
~2 < i0 ⊦ ~1 = i0, abs(i0) < abs(2 + i0)

Rule argument

Additional information

Used simplifier rules: numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;%sub-add: ⊦ i + j - i = j;%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2