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Proof step 4 from proof addarrays-at-other (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# refs = i→n(countrefs(ints)), posints(ints), 0 ≤ i ⊦ (addarrays(refs, ty, ints, i, st)[r -- sk]) = st[r -- sk], ints = [], n→i(# refs) = countrefs(ints), r ∈ refs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]