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Proof step 19 from proof newrefs-new-01 (previous step)

rk = r0 -- sk0, rk ∈ st, r0 ∈ newref_list(m, st[newref(st) -- 0 ', noval]), ∀ r. r ∈ newref_list(m, st[newref(st) -- 0 ', noval]) → r ≠ newref(st) ∧ r ≠ jvmref ∧ (∀ rk. rk ∈ st → rk .ref ≠ r), ∀ st. is_newref_list(newref_list(m, st), st) ⊦ r0 = jvmref, r0 = newref(st), dups(newref_list(m, st[newref(st) -- 0 ', noval])), ∃ sk. r0 -- sk ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
⊦ r -- sk ∈ st → r ∈ st
Proof step 17:
⊦ r ∈ st → newref(st) ≠ r
Proof step 18:
r0 ∈ st, r0 ∈ newref_list(m, st[newref(st) -- 0 ', noval]), r0 -- sk0 ∈ st, ∀ st. is_newref_list(newref_list(m, st), st), ∀ r. r ∈ newref_list(m, st[newref(st) -- 0 ', noval]) → r ≠ newref(st) ∧ r ≠ jvmref ∧ (∀ rk. rk ∈ st → rk .ref ≠ r) ⊦ r0 = jvmref, dups(newref_list(m, st[newref(st) -- 0 ', noval])), ∃ sk. r0 -- sk ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]