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Proof step 7 from proof refs-put (previous step)

r0 ∈ st[r -- sk, val] .refs ⊦ r0 = r ∨ r0 ∈ st .refs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
r0 ∈ st[r -- sk, val] .refs ⊦ st[r -- sk, val] .refs = ∅, r0 = r, # st[r -- sk, val] .refs = 0, r0 ∈ st .refs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ rset ≠ ∅ → # rset ≠ 0], [CSIMPFORWARD : ⊦ r ∈ rset → rset ≠ ∅])]]