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Proof step 5 from proof refs-put (previous step)

r0 ∈ st[r -- sk, val] ⊦ st[r -- sk, val] .refs = ∅, r0 = r, # st[r -- sk, val] .refs = 0, r0 ∈ st .refs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
r0 ∈ st ⊦ st[r -- sk, val] .refs = ∅, r0 = r, # st[r -- sk, val] .refs = 0, r0 ∈ st .refs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ∈ st[r0 -- sk, val] ↔ r = r0 ∨ r ∈ st])]]