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Proof step 2 from proof abs-eq (previous step)

i→n(i) = i→n(j) ⊦ abs(i) = abs(j)

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 3:
⊦ i < j ∨ i = j ∨ j < i
Proof step 4:
abs(i) < abs(j) ∨ abs(j) < abs(i), i→n(i) = i→n(j) ⊦ abs(i) = abs(j)

Rule argument

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

Additional information

Used terms: abs(i), abs(j)