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Proof step 9 from proof count-rec (previous step)

i1 + i = i2 + 1, 0 ≤ i1, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → i0 ≤ j0 → count-i2j(i0, j0) = i0 ' + count-i2j(i0 + 1, j0) ⊦ count-i2j(i, i2) + (i1 + i)' = i ' + count-i2j(i + 1, i1 + i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
i1 + i = i2 + 1, 0 ≤ i1, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → i0 ≤ j0 → count-i2j(i0, j0) = i0 ' + count-i2j(i0 + 1, j0) ⊦ count-i2j(i, i2) + (i2 + 1)' = i ' + count-i2j(i + 1, i2 + 1)

Proof step 9 from proof count-rec (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information