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Proof step 14 from proof le-pos (previous step)

~2 < i0, 1 < i, 2 ≤ i, 0 ≤ i0, ~1 ≤ i0, j0 ≤ 0, j0 ≤ i0, ∀ j1, i, j0. i→n(j1) < i→n(i0 +1) → 1 < i ∧ 2 ≤ i ∧ 0 ≤ j1 → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ i ^ i0 * i ≤ i ^ j0 ↔ i0 + 1 ≤ j0

User applied rule cut formula

And obtained the following new goals

Proof step 15:
j0 = 0, 1 < i, ~2 < i0, j0 ≤ i0, j0 ≤ 0, ~1 ≤ i0, 0 ≤ i0, 2 ≤ i, ∀ j1, i, j0. i→n(j1) < i→n(i0 +1) → 1 < i ∧ 2 ≤ i ∧ 0 ≤ j1 → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ i ^ i0 * i ≤ i ^ j0 ↔ i0 + 1 ≤ j0
Proof step 21:
1 < i, ~2 < i0, j0 ≤ i0, j0 ≤ 0, ~1 ≤ i0, 0 ≤ i0, 2 ≤ i, ∀ j1, i, j0. i→n(j1) < i→n(i0 +1) → 1 < i ∧ 2 ≤ i ∧ 0 ≤ j1 → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ i ^ i0 * i ≤ i ^ j0 ↔ i0 + 1 ≤ j0, j0 = 0

Rule argument

fma: j0 = 0

Additional information

Used formulas: j0 = 0