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Proof step 5 from proof newref-03 (conclusion)

⊦ r ∈ st → ¬ newref(r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ r ∈ st ↔ (∃ sk. r -- sk ∈ st)
Proof step 2:
⊦ newref(r, st) ↔ r ≠ jvmref ∧ (∀ rk. rk ∈ st → rk .ref ≠ r)
Proof step 3:
⊦ r -- sk ∈ st → r ∈ st
Proof step 4:
r -- sk ∈ st, r -- sk0 ∈ st, ∀ rk. rk ∈ st → rk .ref ≠ r ⊦ r = jvmref

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]