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Proof step 5 from proof new-out (previous step)

rk = r -- sk, rk ∈ st, ∀ rk0. rk0 = _out ∨ rk0 ∈ st → rk0 .ref ≠ r ⊦ r = jvmref

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ r -- sk ∈ st → r ∈ st
Proof step 4:
r -- sk ∈ st, r ∈ st, ∀ rk0. rk0 = _out ∨ rk0 ∈ st → rk0 .ref ≠ r ⊦ r = jvmref

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]