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Proof step 5 from proof count-i2j-ordered (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i ≤ j → count-i2j(i, j) = i ' + count-i2j(i + 1, j)
Proof step 7:
¬ i ≤ i1 + i, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → true → ordered≤(count-i2j(i0, j0)) ⊦ ordered≤(count-i2j(i, i1 + i))
Proof step 10:
i ≤ i1 + i, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → true → ordered≤(count-i2j(i0, j0)) ⊦ ordered≤(i ' + count-i2j(i + 1, i1 + i))

Rule argument

spec: inst: name: count-rec seq: ⊦ i ≤ j → count-i2j(i, j) = i ' + count-i2j(i + 1, j) subst: [j, i] → [i1 + i, i] flags: rewrite, current

Additional information

Used terms: i1 + i, i