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Proof step 11 from proof newref-newreflist-subset (previous step)

st0 ⊆ st, r ∈ newref_list(m, st[newref(st) -- 0 ', noval]), r ∈ newref_list(m, st[newref(st) -- 0 ', noval]) ∧ st0 ⊆ st[newref(st) -- 0 ', noval] → r ≠ jvmref ∧ (∀ rk. rk ∈ st0 → rk .ref ≠ r) ⊦ r = newref(st), newref(r, st0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
st0 ⊆ st, r ∈ newref_list(m, st[newref(st) -- 0 ', noval]), st0 ⊆ st[newref(st) -- 0 ', noval] → r ≠ jvmref ∧ (∀ rk. rk ∈ st0 → rk .ref ≠ r) ⊦ r = newref(st), newref(r, st0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]