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Proof step 14 from proof refs-put (previous step)

⊦ ∀ r0. r0 ∈ st[r -- sk, val] .refs ↔ r0 ∈ st .refs ++ r

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
⊦ r0 ∈ st[r -- sk, val] .refs ↔ r0 = r ∨ r0 ∈ st .refs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ∈ rset ++ r0 ↔ r = r0 ∨ r ∈ rset])]]