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Proof step 6 from proof split-init-2 (previous step)

split([], A, X), partitions([], # A), ∀ p. p ∈ [] → p.count = 0 ⊦ split([], A, [])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
split([], A, X), partitions([], # A) ⊦ split([], A, [])

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(list,partition,in-empty, ⊦ ¬ p ∈ [])))]