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Proof step 10 from proof refs-del (previous step)

⊦ ∀ r0. r0 ∈ (st -- r) .refs ↔ r0 ∈ st .refs -- r

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ r0 ∈ (st -- r) .refs ↔ r0 ≠ r ∧ r0 ∈ st .refs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ∈ rset -- r0 ↔ ¬ ¬ (r ≠ r0 ∧ r ∈ rset)])]]