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Proof step 11 from proof addarrays-subset (previous step)

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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# refs = i→n(countrefs(ints)), posints(ints), r -- sk ∈ st, is_newref_list(refs, st), 0 ≤ i, r ∈ st, r ∈ refs, 0 ≤ countrefs(ints) ⊦ st[r -- sk] = (addarrays(refs, ty, ints, i, st)[r -- sk]), ints = []

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j])]]