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Proof step 24 from proof addarrays-refin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
n→i(# refs) = countrefs(ints), posints(ints) ⊦ ints = [], r -- sk ∈ addarrays(refs, ty, ints, i, st) ↔ r -- sk ∈ st, r ∈ refs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]