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Proof step 8 from proof abs-sub-01 (previous step)

~2 < i0 ⊦ ~1 = i0, abs(i0) < abs(2 + i0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ ¬ j < 0 → abs(j) = j
Proof step 10:
~2 < i0 ⊦ ~1 = i0, abs(i0) < i0 + 2, i0 + 2 < 0

Rule argument

spec: <toplevel> name: Abs-pos seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [i0 + 2] flags: <all paths>

Additional information

Used simplifier rules: numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ i < j → ¬ j < i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: i0 + 2