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Proof step 5 from proof newref-02 (previous step)

∀ rk. rk = r0 -- sk ∨ rk ∈ st → rk .ref ≠ r ⊦ r -- sk0 ∈ st → r1 ≠ 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 -- sk0 ∈ st, r ∈ st, ∀ rk. rk = r0 -- sk ∨ rk ∈ st → rk .ref ≠ r ⊦ r = jvmref

Rule argument

[emptyarg]

Additional information

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