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

# 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)) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ 0 ≤ i → abs((* i)) = i
Proof step 7:
⊦ ¬ countrefs(ints) < 0
Proof step 8:
⊦ 0 < i → ¬ countrefs(i ' + ints) < countrefs(ints)

Rule argument

[speclemmaarg: ,0 ≤ i → abs((* i)) = i ,[i] → [countrefs(ints)] ,F ,int-abs , ,numabspos ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1])], [EXTRATERMS : (countrefs(ints))]]