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Proof step 11 from proof addarrays-at-other-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ abs(countrefs(ints)) = countrefs(ints)
Proof step 8:
⊦ countrefs([]) = 1
Proof step 10:
i < 0, r ∈ st ⊦ (addarrays(newref(st) ', ty, [], i, st)[r -- sk]) = st[r -- sk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ posints([])], [CSIMPSEQ : ⊦ newref_list(1, st) = newref(st) '], [CSIMPSEQ : ⊦ n→i(i→n(i)) = abs(i)], [CSIMPSEQ : ⊦ # newref_list(n, st) = n])]]