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Proof step 23 from proof eqs-step (conclusion)

⊦ eqS(sys, sys0) ∧ eqd(sys, cdom(co), sys0) → eqS(step(sys, co), step(sys0, co))

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : eqd-symmetry, , , ⊦ eqd(sys, d, sys0) → eqd(sys0, d, sys)], [CSIMPNAMEDSEQ : eqd-transitivity, , , ⊦ eqd(sys, d, sys0) ∧ eqd(sys0, d, sys1) → eqd(sys, d, sys1)], [CSIMPNAMEDSEQ : eqd-reflexivity, , , ⊦ eqd(sys, d, sys)])]]