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Proof step 12 from proof transex-enabled-intrans (conclusion)

⊦ trans∃(trans, S0, tr, s0) ∧ enabled(trans, s0, tr1) → (¬ tr + tr1 ∈TR trans ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]