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Proof step 14 from proof countrefs-le (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ i ≤ 0 ∧ j ≤ 0 → (abs(i) < abs(j) ↔ j < i)
Proof step 12:
¬ (countrefs(i ' + ints) ≤ 0 ∧ countrefs(ints) ≤ 0), # refs = i→n(countrefs(i ' + ints)), # refs < i→n(countrefs(ints)), 0 < i, # refs ≤ # refs - i→n(countrefs(ints)), abs(countrefs(i ' + ints)) < abs(countrefs(ints)) ⊦
Proof step 13:
⊦ ¬ countrefs(ints) < 0

Rule argument

[speclemmaarg: ,i ≤ 0 ∧ j ≤ 0 → (abs(i) < abs(j) ↔ j < i) ,[j, i] → [countrefs(ints), countrefs(i ' + ints)] ,F ,int-abs , ,absls-02 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i < j ∧ j ≤ j0 → i < j0], [CSIMPFORWARD : ⊦ i ≤ j ∧ j < j0 → i < j0], [CSIMPFORWARD : ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0])], [EXTRATERMS : (countrefs(ints), countrefs(i ' + ints))]]