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Proof step 3 from proof countrefs2-zero (conclusion)

⊦ countrefs2(0 ' + ints) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

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

Rule argument

[crewrite-arg: countrefs_more_le0, F, [ints] → [0 ' + ints]]

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 + 1 ≠ 0], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ 1 = m + 1 ↔ m = 0], [CSIMPSEQ : ⊦ # ints = 0 ↔ ints = []], [CSIMPSEQ : ⊦ (i ' + ints) .first = i], [CSIMPSEQ : ⊦ ints + [] = ints])], [EXTRATERMS : (0 ' + ints)]]