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Proof step 5 from proof union-empty-01 (previous step)

⊦ ∀ rk. (rk ∈ ∅ ∪ st ↔ rk ∈ st) ∧ (∅ ∪ st)[rk] = st[rk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ rk ∈ st ∪ st0 ↔ rk ∈ st ∨ rk ∈ st0
Proof step 4:
⊦ (∅ ∪ st)[rk] = st[rk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ rk ∈ ∅])]]