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Proof step 9 from proof countrefs-not-zero (previous step)

¬ # (i ' + ints0) ≤ 1, 0 ≤ countrefs(ints0), countrefs(i ' + ints0) < 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 ≤ countrefs(ints0), countrefs(i ' + ints0) < 0 ⊦ ints0 = []

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ # ints = 0 ↔ ints = []], [CSIMPSEQ : ⊦ m + n ≤ m ↔ n = 0], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ # i ' = 0 +1], [CSIMPSEQ : ⊦ # (ints + ints0) = # ints + # ints0])]]