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Proof step 2 from proof abssub-05 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
j < i0 ⊦ abs(i - j) + j = i, i + 1 < i0

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;