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Proof step 5 from proof instore-two (conclusion)

⊦ r -- (sk ' + sk0 ') ∈ st ↔ r -- sk ∈ st ∧ st[r -- sk] .refval -- sk0 ' ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ r -- sk ' ∈ st ↔ r -- sk ∈ st
Proof step 4:
⊦ r -- (sk ' + sk0 ') ∈ st ↔ r -- sk ∈ st ∧ st[r -- sk] .refval -- sk0 ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]