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Proof step 6 from proof abs-eq (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ 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: ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used terms: i, j