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Proof step 14 from proof union-in-01 (previous step)

r -- sk ∈ st ∨ r -- sk0 ∈ st0 ⊦ r -- sk ∈ st ∨ r -- sk ∈ st0, ∃ sk1. r -- sk1 ∈ st ∨ r -- sk1 ∈ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

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