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Proof step 11 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)) ⊦ ¬ 0 < countrefs(i ' + ints) ∧ ¬ 0 < countrefs(ints)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ i → abs(i) = i
Proof step 4:
⊦ ¬ countrefs(ints) < 0
Proof step 10:
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)

Rule argument

[speclemmaarg: ,0 ≤ i → abs(i) = i ,[i] → [countrefs(i ' + ints)] ,F ,int-abs , ,absle ]

Additional information

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