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Proof step 9 from proof countrefs-not-zero-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
countrefs(i ' + ints0) = 0 ⊦ countrefs(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])]]