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Proof step 4 from proof instore-inrefs-02 (conclusion)

⊦ ¬ r ∈ st → ¬ r ∈ st .refs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ st .refs = ∅ ↔ st = ∅
Proof step 3:
r ∈ st .refs ⊦ st = ∅, r ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r ∈ rset → rset ≠ ∅])]]