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Proof step 5 from proof newref-put-in (previous step)

¬ (∀ r. r ∈ st[rk, val] ↔ r ∈ st) ⊦ rk ∈ st → newref(st[rk, val]) = newref(st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ r ∈ st[rk, val] ↔ r = rk .ref ∨ r ∈ st
Proof step 4:
rk ∈ st ⊦ newref(st[rk, val]) = newref(st), rk .ref ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]