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Proof step 12 from proof abs-eq (previous step)

abs(i) < abs(j) ∨ i→n(j) < i→n(i), abs(i) = abs(j) ⊦ i→n(i) = i→n(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ abs(i) < abs(j) ↔ i→n(i) < i→n(j)

Rule argument

spec: intnat inst: name: rlabs seq: ⊦ abs(i) < abs(j) ↔ i→n(i) < i→n(j) subst: [j, i] → [i, j] flags: rewrite, current

Additional information

Used simplifier rules: irref: ⊦ ¬ i < i; from specification int-basic1 Used terms: i, j