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Proof step 12 from proof ls-pos (previous step)

j < j0, 0 < i, 1 < i, 0 < j0, 0 ≤ j, i ^ j0 ≤ i ^ j ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ 1 < i ∧ 0 ≤ j → (i ^ j ≤ i ^ j0 ↔ j ≤ j0)

Rule argument

spec: <toplevel> name: le-pos seq: ⊦ 1 < i ∧ 0 ≤ j → (i ^ j ≤ i ^ j0 ↔ j ≤ j0) subst: [j0, j, i] → [j, j0, i] flags: <all paths>

Additional information

Used simplifier rules: ⊦ i ≤ j → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s) Used terms: j, j0, i