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Proof step 11 from proof count-i2j-ordered (previous step)

0 ≤ i1, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → true → ordered≤(count-i2j(i0, j0)) ⊦ ordered≤(i ' + count-i2j(i + 1, i1 + i))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ ints ≠ [] ∧ ints.rest ≠ [] → (ordered≤(ints) ↔ ¬ ints.rest.first < ints.first ∧ ordered≤(ints.rest))
Proof step 13:
¬(i ' + count-i2j(i + 1, i1 + i) ≠ [] ∧ (i ' + count-i2j(i + 1, i1 + i)).rest ≠ []), 0 ≤ i1, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → true → ordered≤(count-i2j(i0, j0)) ⊦ ordered≤(i ' + count-i2j(i + 1, i1 + i))
Proof step 16:
i ' + count-i2j(i + 1, i1 + i) ≠ [] ∧ (i ' + count-i2j(i + 1, i1 + i)).rest ≠ [], 0 ≤ i1, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → true → ordered≤(count-i2j(i0, j0)) ⊦ ¬ (i ' + count-i2j(i + 1, i1 + i)).rest.first < (i ' + count-i2j(i + 1, i1 + i)).first ∧ ordered≤((i ' + count-i2j(i + 1, i1 + i)).rest)

Rule argument

spec: olist inst: intlista name: ole-rec seq: ⊦ ints ≠ [] ∧ ints.rest ≠ [] → ( ordered≤(ints) ↔ ¬ ints.rest.first < ints.first ∧ ordered≤(ints.rest)) subst: [ints] → [i ' + count-i2j(i + 1, i1 + i)] flags: rewrite

Additional information

Used terms: i ' + count-i2j(i + 1, i1 + i)