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Proof step 12 from proof newref-newreflist-subset (previous step)

st0 ⊆ st, r ∈ newref_list(m, st[newref(st) -- 0 ', noval]), ∀ r0, st1, st2. r0 ∈ newref_list(m, st1) ∧ st2 ⊆ st1 → r0 ≠ jvmref ∧ (∀ rk. rk ∈ st2 → rk .ref ≠ r0) ⊦ r = newref(st), newref(r, st0)

User applied rule all left

And obtained the following new goals

Proof step 11:
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)

Rule argument

[EXRARG : pos: left 3, [extquanttermlist: (r, st[newref(st) -- 0 ', noval], st0) abort:F , discard:T , computed:F ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r, st[newref(st) -- 0 ', noval], st0)]]