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Proof step 4 from proof new-insert (previous step)

rk0 = r -- sk, rk0 ∈ st, ∀ rk1. rk1 = rk ∨ rk1 ∈ st → rk1 .ref ≠ r ⊦ r = jvmref

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ r -- sk ∈ st → r ∈ st
Proof step 3:
r -- sk ∈ st, r ∈ st, ∀ rk1. rk1 = rk ∨ rk1 ∈ st → rk1 .ref ≠ r ⊦ r = jvmref

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]