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Proof step 5 from proof ls-pos (previous step)

i ^ j < i ^ j0, 0 < i, 1 < i, 0 ≤ j, j0 ≤ j ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 6:
⊦ 1 < i ∧ 0 ≤ j → (i ^ j ≤ i ^ j0 ↔ j ≤ j0)
Proof step 7:
1 < i, 0 < i, i ^ j < i ^ j0, j0 ≤ j, 0 ≤ j, 1 < i ∧ 0 ≤ j → (i ^ j ≤ i ^ j0 ↔ j ≤ j0) ⊦

Rule argument

spec: <toplevel> name: le-pos seq: ⊦ 1 < i ∧ 0 ≤ j → (i ^ j ≤ i ^ j0 ↔ j ≤ j0) subst: [i, j, j0] → [i, j, j0]

Additional information

Used terms: i, j, j0