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Proof step 10 from proof countrefs-le (previous step)

0 ≤ countrefs(i ' + ints), # refs = i→n(countrefs(i ' + ints)), # refs < i→n(countrefs(ints)), 0 < i, # refs ≤ # refs - i→n(countrefs(ints)), countrefs(i ' + ints) < abs(countrefs(ints)) ⊦ ¬ 0 < countrefs(i ' + ints) ∧ ¬ 0 < countrefs(ints)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ countrefs(ints) ≠ 0
Proof step 9:
# refs = i→n(countrefs(i ' + ints)), 0 < countrefs(i ' + ints), 1 < abs(countrefs(ints)), 0 ≤ countrefs(i ' + ints), countrefs(i ' + ints) < abs(countrefs(ints)), # refs ≤ # refs - i→n(countrefs(ints)), 0 < i, # refs < i→n(countrefs(ints)) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 0 < abs(i) ↔ i ≠ 0], [CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPFORWARD : ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j], [CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPFORWARD : ⊦ i < i0 ∧ i0 < i1 → i < i1], [CSIMPSEQ : ⊦ i ≤ j → (j ≤ i ↔ i = j)], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i])]]