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Proof step 23 from proof del-switch (previous step)

¬ r1 -- sk ∈ st -- r0 -- r1 ⊦ r0 = r1, (st -- r1)[r1 -- sk] = ∅[r1 -- sk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
⊦ r -- sk ∈ st -- r0 ↔ ¬ ¬ (r ≠ r0 ∧ r -- sk ∈ st)
Proof step 22:
⊦ r0 = r1, (st -- r1)[r1 -- sk] = ∅[r1 -- sk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]