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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: rlabs seq: ⊦ abs(i) < abs(j) ↔ i→n(i) < i→n(j) subst: [j, i] → [i, j] flags: <rotated> <all paths>

Additional information

Used simplifier rules: ⊦ ¬ i < i; (s) Used terms: i, j