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Proof step 8 from proof le-pos-01 (previous step)

0 < j, i0 < i, 0 < i, 0 ≤ i, 0 ≤ i0, 0 ≤ j, i ^ j ≤ i0 ^ j, ∀ j0, i, i0. i→n(j0) < i→n(j) → 0 < i ∧ i0 < i ∧ 0 < j0 ∧ i ^ j0 ≤ i0 ^ j0 ∧ 0 ≤ j0 ∧ 0 ≤ i0 ∧ 0 ≤ i → false ⊦

User applied rule apply induction

And obtained the following new goals

Proof step 9:
0 < i, i0 < i, 0 < j, i ^ j ≤ i0 ^ j, 0 ≤ j, 0 ≤ i0, 0 ≤ i, ∀ j0, i, i0. i→n(j0) < i→n(j) → 0 < i ∧ i0 < i ∧ 0 < j0 ∧ i ^ j0 ≤ i0 ^ j0 ∧ 0 ≤ j0 ∧ 0 ≤ i0 ∧ 0 ≤ i → false ⊦ 0 < i ∧ i0 < i ∧ 0 < j - 1 ∧ i ^ (j - 1) ≤ i0 ^ (j - 1) ∧ 0 ≤ j - 1 ∧ 0 ≤ i0 ∧ 0 ≤ i

Rule argument

subst: [j0, i, i0] → [j - 1, i, i0]

Additional information

Used simplifier rules: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s) Used terms: j - 1, i, i0