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Proof step 4 from proof abs-eq (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ abs(i) < abs(j) ↔ i→n(i) < i→n(j)
Proof step 6:
i→n(i) = i→n(j), i→n(i) < i→n(j) ∨ abs(j) < abs(i) ⊦ abs(i) = abs(j)

Rule argument

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

Additional information

Used terms: i, j