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Proof step 9 from proof addarrays-subset (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
# refs = i→n(countrefs(ints)), 0 ≤ countrefs(ints), r ∈ refs, r ∈ st, 0 ≤ i, r -- sk ∈ st, posints(ints), ∀ r. r ∈ refs → newref(r, st) ⊦ ints = [], st[r -- sk] = (addarrays(refs, ty, ints, i, st)[r -- sk]), dups(refs)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]