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Proof step 9 from proof refs-empty-01 (previous step)

∀ r. ¬ r ∈ st .refs ⊦ ∀ rk. (rk ∈ st ↔ rk ∈ ∅) ∧ st[rk] = ∅[rk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ ∅[rk] = noval
Proof step 8:
rk ∈ st, ∀ r. ¬ r ∈ st .refs ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ rk ∈ st → st[rk] = ∅[rk]], [CSIMPSEQ : ⊦ ¬ rk ∈ ∅])]]