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Proof step 9 from proof countrefs-ls (previous step)

# (i ' + ints) > 1 ∧ (i ' + ints) .first > 0, 1 + (i ' + ints) .first * countrefs((i ' + ints) .rest) < countrefs(ints), 0 < i ⊦ ints = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 < i, 1 + countrefs(ints) * i < countrefs(ints) ⊦ ints = []

Rule argument

[emptyarg]

Additional information

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