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Proof step 2 from proof labs-01 (previous step)

⊦ i→n(i) ≤ i→n(j) ↔ abs(i) ≤ abs(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ i ≤ j ↔ ¬ j < i
Proof step 4:
⊦ i→n(i) ≤ i→n(j) ↔ ¬ abs(j) < abs(i)

Rule argument

spec: int-abs inst: name: le-def seq: ⊦ i ≤ j ↔ ¬ j < i subst: [j, i] → [abs(j), abs(i)] flags: rewrite

Additional information

Used terms: abs(j), abs(i)