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Proof step 6 from proof countrefs-not-zero-01 (previous step)

countrefs(i ' + ints0) = 0, 0 < i ⊦ ints0 = [], countrefs(ints0) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ # ints > 1 ∧ ints .first > 0 → countrefs(ints) = 1 + ints .first * countrefs(ints .rest)
Proof step 5:
⊦ ¬ countrefs(ints) < 0

Rule argument

[crewrite-arg: countrefs-rec, F, [ints] → [i ' + ints0]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i < j → (¬ ¬ j ≤ i ↔ false)], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i], [CSIMPSEQ : ⊦ 1 = m + 1 ↔ m = 0], [CSIMPSEQ : ⊦ (i ' + ints) .rest = ints], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ (i ' + ints) .first = i], [CSIMPSEQ : ⊦ ((* j)) = i0 + ((* j0)) ↔ j - j0 = i0], [CSIMPSEQ : ⊦ i * j = (~1) ↔ abs(i) = 1 ∧ i = ~ j], [CSIMPSEQ : ⊦ ~ i = j ↔ i + j = 0], [CSIMPSEQ : ⊦ i * j * j0 = (i * j) * j0], [CSIMPSEQ : ⊦ i * j = j * i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ # (ints + ints0) = # ints + # ints0], [CSIMPSEQ : ⊦ # i ' = 0 +1], [CSIMPSEQ : ⊦ m > n ↔ n < m], [CSIMPSEQ : ⊦ 1 < m ↔ ¬ (m = 0 ∨ m = 1)], [CSIMPSEQ : ⊦ m + 1 ≠ 0], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1], [CSIMPSEQ : ⊦ # ints = 0 ↔ ints = []], [CSIMPSEQ : ⊦ i > j ↔ j < i], [CSIMPFORWARD : ⊦ i < i0 ∧ i = i1 → i1 < i0], [CSIMPSEQ : ⊦ i + j < j ↔ i < 0])], [EXTRATERMS : (i ' + ints0)]]