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Proof step 11 from proof countrefs-rec-02 (previous step)

0 ≤ i2 ⊦ i→n(countrefs(i2 ' + i1 ' + ints1)) = i→n(countrefs(i1 ' + ints1)) * i→n(i2) + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ # ints > 1 ∧ ints .first > 0 → countrefs(ints) = 1 + ints .first * countrefs(ints .rest)
Proof step 2:
⊦ countrefs(0 ' + ints) = 1
Proof step 10:
# (i2 ' + i1 ' + ints1) > 1 ∧ (i2 ' + i1 ' + ints1) .first > 0, 0 ≤ i2 ⊦ i→n(1 + (i2 ' + i1 ' + ints1) .first * countrefs((i2 ' + i1 ' + ints1) .rest)) = i→n(countrefs(i1 ' + ints1)) * i→n(i2) + 1

Rule argument

[crewrite-arg: countrefs-rec, F, [ints] → [i2 ' + i1 ' + ints1]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # (ints + ints0) = # ints + # ints0], [CSIMPSEQ : ⊦ # i ' = 0 +1], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ m > n ↔ n < m], [CSIMPSEQ : ⊦ 1 < m ↔ ¬ (m = 0 ∨ m = 1)], [CSIMPSEQ : ⊦ m + n = 0 ↔ m = 0 ∧ n = 0], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : n < n0 ⊦ ((* n)) ≠ m0 + ((* n0))], [CSIMPSEQ : ⊦ (i ' + ints) .first = i], [CSIMPSEQ : ⊦ i > j ↔ j < i], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i], [CSIMPSEQ : ⊦ i ≤ j → (j ≤ i ↔ i = j)], [CSIMPSEQ : ⊦ m * 0 = 0], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n], [CSIMPSEQ : ⊦ m * n = n * m])], [EXTRATERMS : (i2 ' + i1 ' + ints1)]]