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Proof step 8 from proof union-in-01 (previous step)

r -- sk ∈ st0, r ∈ st ∪ st0 ⊦ ∃ sk. r -- sk ∈ st, ∃ sk. r -- sk ∈ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st])]]