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Proof step 7 from proof countrefs-not-zero-01 (previous step)

¬ (# (i ' + ints0) > 1 ∧ (i ' + ints0) .first ≤ 0), countrefs(i ' + ints0) = 0 ⊦ countrefs(ints0) = 0, ints0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
countrefs(i ' + ints0) = 0, 0 < i ⊦ ints0 = [], countrefs(ints0) = 0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPFORWARD : ⊦ i < i0 ∧ i = i1 → i1 < i0], [CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i], [CSIMPSEQ : ⊦ (i ' + ints) .first = i], [CSIMPSEQ : ⊦ # ints = 0 ↔ ints = []], [CSIMPSEQ : ⊦ 1 = m + 1 ↔ m = 0], [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])]]