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Proof step 8 from proof abs-eq (previous step)

abs(i) = abs(j) ⊦ i→n(i) = i→n(j)

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 9:
⊦ n < m ∨ n = m ∨ m < n
Proof step 10:
i→n(i) < i→n(j) ∨ i→n(j) < i→n(i), abs(i) = abs(j) ⊦ i→n(i) = i→n(j)

Rule argument

spec: nat-basic1 inst: name: tot seq: ⊦ n < m ∨ n = m ∨ m < n subst: [n, m] → [i→n(i), i→n(j)] flags:

Additional information

Used terms: i→n(i), i→n(j)