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Proof step 3 from proof le-pos-01 (previous step)

0 < j, 0 ≤ i0, 0 ≤ i ⊦ i ^ j ≤ i0 ^ j ↔ i ≤ i0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i ≤ i0 ∧ 0 ≤ i ∧ 0 ≤ j → i ^ j ≤ i0 ^ j
Proof step 5:
0 < j, 0 ≤ i, 0 ≤ i0 ⊦ i ^ j ≤ i0 ^ j ↔ i ≤ i0, i ≤ i0 ∧ 0 ≤ i ∧ 0 ≤ j

Rule argument

spec: <toplevel> name: ls-pos-01 seq: ⊦ i ≤ i0 ∧ 0 ≤ i ∧ 0 ≤ j → i ^ j ≤ i0 ^ j subst: [i0, j, i] → [i0, j, i] flags: <all paths>

Additional information

Used terms: i0, j, i