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Proof step 5 from proof ls-pos-02 (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 6:
i0 ^ j < i ^ j, i ≤ i0, 0 ≤ j, 0 ≤ i, 0 ≤ i0, ∀ j0, i0, i. i→n(j0) < i→n(j) → i0 ^ j0 < i ^ j0 ∧ i ≤ i0 ∧ 0 ≤ j0 ∧ 0 ≤ i ∧ 0 ≤ i0 → false ⊦ i0 ^ (j - 1) < i ^ (j - 1) ∧ i ≤ i0 ∧ 0 ≤ j - 1 ∧ 0 ≤ i ∧ 0 ≤ i0

Rule argument

subst: [j0, i0, i] → [j - 1, i0, i]

Additional information

Used simplifier rules: Pot-zero: ⊦ j ^ 0 = 1;% ⊦ 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, i0, i