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Proof step 3 from proof i2i-pos (previous step)

i→n(n→i(i→n(i))) = i→n(i), 0 < i ⊦ n→i(i→n(i)) = i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i→n(i) = i→n(j) ↔ abs(i) = abs(j)

Rule argument

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

Additional information

Used simplifier rules: abs-02: ⊦ abs(n→i(n)) = n→i(n);%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs Used terms: i, n→i(i→n(i))