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Proof step 17 from proof addarrays-subset (previous step)

r -- sk ∈ st, is_newref_list(refs, st), 0 ≤ i ⊦ st[r -- sk] = (addarrays(refs, ty, ints, i, st)[r -- sk])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
r -- sk ∈ st, is_newref_list(refs, st), 0 ≤ i, r ∈ st ⊦ st[r -- sk] = (addarrays(refs, ty, ints, i, st)[r -- sk])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st])]]