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Proof step 3 from proof union-in-02 (conclusion)

⊦ ¬ r ∈ st .refs → (r -- sk ∈ st ∪ st0 ↔ r -- sk ∈ st0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ rk ∈ st ∪ st0 ↔ rk ∈ st ∨ rk ∈ st0
Proof step 2:
⊦ ¬ r ∈ st .refs → ¬ r -- sk ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]