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Proof step 4 from proof le-pos (previous step)

2 ≤ i, 0 ≤ j, 1 < i, ∀ j1, i, j0. i→n(j1) < i→n(j) → 2 ≤ i ∧ 0 ≤ j1 ∧ 1 < i → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ i ^ j ≤ i ^ j0 ↔ j ≤ j0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < i → j ^ i = j ^ i -1 * j
Proof step 6:
¬ 0 < j, 2 ≤ i, 0 ≤ j, 1 < i, ∀ j1, i, j0. i→n(j1) < i→n(j) → 2 ≤ i ∧ 0 ≤ j1 ∧ 1 < i → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ i ^ j ≤ i ^ j0 ↔ j ≤ j0
Proof step 8:
0 < j, 2 ≤ i, 0 ≤ j, 1 < i, ∀ j1, i, j0. i→n(j1) < i→n(j) → 2 ≤ i ∧ 0 ≤ j1 ∧ 1 < i → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ i ^ j -1 * i ≤ i ^ j0 ↔ j ≤ j0

Rule argument

spec: int-pot inst: name: Pot-pos seq: ⊦ 0 < i → j ^ i = j ^ i -1 * j subst: [i, j] → [j, i] flags: rewrite, current

Additional information

Used terms: j, i