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Proof step 3 from proof countrefs2-two (conclusion)

0 < i ⊦ countrefs2(i ' + j ') = i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ # ints > 1 ∧ ints .first > 0 → countrefs2(ints) = ints .first + ints .first * countrefs2(ints .rest)
Proof step 2:
⊦ countrefs2(i ') = 0

Rule argument

[crewrite-arg: countrefs_more_gr0, F, [ints] → [i ' + j ']]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i < j → (¬ ¬ j ≤ i ↔ false)], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i], [CSIMPSEQ : ⊦ i > j ↔ j < i], [CSIMPSEQ : ⊦ # i ' = 0 +1], [CSIMPSEQ : ⊦ # (ints + ints0) = # ints + # ints0], [CSIMPSEQ : ⊦ (i ' + ints) .rest = ints], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ (i ' + ints) .first = i], [CSIMPSEQ : ⊦ i * 0 = 0], [CSIMPSEQ : ⊦ i * j * j0 = (i * j) * j0], [CSIMPSEQ : ⊦ i * j = j * i], [CSIMPSEQ : ⊦ i + 0 = i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i])], [EXTRATERMS : (i ' + j ')]]