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Proof step 15 from proof del-put-switch (previous step)

¬ r1 -- sk0 ∈ st[r -- sk, val] -- r1 ⊦ r = r1, ∅[r1 -- sk0] = (st -- r1)[r1 -- sk0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
⊦ r -- sk ∈ st -- r0 ↔ ¬ ¬ (r ≠ r0 ∧ r -- sk ∈ st)
Proof step 14:
⊦ r = r1, ∅[r1 -- sk0] = (st -- r1)[r1 -- sk0]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]