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Proof step 3 from proof fac-add (previous step)

0 ≤ i ∧ 0 ≤ j, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 0 ≤ j0 → fac(i + j0) = fac(i) * Π countFun(1, j0, λ j0. i + j0) ⊦ fac(i + j) = fac(i) * Π countFun(1, j, λ j0. i + j0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 ≤ j, 0 ≤ i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 0 ≤ j0 → fac(i + j0) = fac(i) * Π countFun(1, j0, λ j0. i + j0) ⊦ fac(i + j) = fac(i) * Π countFun(1, j, λ j0. i + j0)

Proof step 3 from proof fac-add (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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