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Proof step 5 from proof newref-in-01 (previous step)

∃ sk. r0 -- sk ∈ st, ∀ rk. rk ∈ st → rk .ref ≠ r0 ⊦ r0 = jvmref

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ r -- sk ∈ st → r ∈ st
Proof step 4:
r0 -- sk ∈ st, r0 ∈ st, ∀ rk. rk ∈ st → rk .ref ≠ r0 ⊦ r0 = jvmref

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]