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Proof step 11 from proof del-refs-size (previous step)

∃ sk. r -- sk ∈ st, # st ≤ # (st -- r) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ r -- sk ∈ st → r ∈ st
Proof step 10:
r -- sk ∈ st, # st ≤ # (st -- r), r ∈ st ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]