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Proof step 16 from proof eqds-trans (conclusion)

⊦ eqds(sys, ds, sys0) ∧ eqds(sys0, ds, sys1) → eqds(sys, ds, sys1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
eqds(sys0, ds, sys1), eqds(sys, ds, sys0) ⊦ eqds(sys, ds, sys1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]