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Proof step 3 from proof two-ls (previous step)

0 < j, j < 2 ⊦ j = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i < j ↔ i +1 = j ∨ i +1 < j
Proof step 5:
0 < j, j +1 = 2 ∨ j +1 < 2 ⊦ j = 1

Rule argument

spec: <toplevel> name: lsrec seq: ⊦ i < j ↔ i +1 = j ∨ i +1 < j subst: [j, i] → [2, j] flags: <all paths>

Additional information

Used terms: 2, j