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Proof step 8 from proof newrefs-new (previous step)

rk = r -- sk, rk ∈ st, r ∈ newref_list(m, st[newref(st) -- 0 ', noval]), ∀ rk0. rk0 = newref(st) -- 0 ' ∨ rk0 ∈ st → rk0 .ref ≠ r ⊦ r = jvmref, r = newref(st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ r -- sk ∈ st → r ∈ st
Proof step 7:
r -- sk ∈ st, r ∈ newref_list(m, st[newref(st) -- 0 ', noval]), r ∈ st, ∀ rk0. rk0 = newref(st) -- 0 ' ∨ rk0 ∈ st → rk0 .ref ≠ r ⊦ r = jvmref, r = newref(st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]