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Proof step 12 from proof addarrays-subset (previous step)

n→i(# refs) = countrefs(ints), posints(ints), r -- sk ∈ st, is_newref_list(refs, st), 0 ≤ i, r ∈ st ⊦ st[r -- sk] = (addarrays(refs, ty, ints, i, st)[r -- sk]), ints = [], ¬ r ∈ refs ∧ # refs = i→n(countrefs(ints))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
0 ≤ i ⊦ n→i(n) = i ↔ n = i→n(i)
Proof step 6:
⊦ ¬ countrefs(ints) < 0
Proof step 11:
0 ≤ countrefs(ints), # refs = i→n(countrefs(ints)), posints(ints), r -- sk ∈ st, is_newref_list(refs, st), 0 ≤ i, r ∈ st ⊦ st[r -- sk] = (addarrays(refs, ty, ints, i, st)[r -- sk]), ints = [], ¬ r ∈ refs ∧ # refs = i→n(countrefs(ints))

Rule argument

[speclemmaarg: 0 ≤ i ,n→i(n) = ((* i)) ↔ n = i→n(i) ,[i, n] → [countrefs(ints), # refs] ,F ,intnat , ,num-n2i-eq ]

Additional information

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