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Proof step 86 from proof hasSharing-addobj-same (previous step)

is_referencevalue(addobj(r, class, fis, st)[r -- sks]) ∧ sks ≠ [], (addobj(r, class, fis, st)[r -- sks0]) = refval(r0), fis .keys = instfields(class), (addobj(r, class, fis, st)[(addobj(r, class, fis, st)[r -- sks .first]) .refval -- sks .rest]) = refval(r0), disjoint(str, fis .vals .nonnullrefs, st), fits(r, sks, addobj(r, class, fis, st)), fits(r, sks0, addobj(r, class, fis, st)), newref(r, st), validfieldinits(str, fis, st) ⊦ sks0 = sks, sks0 = [], sks = [], r0 = jvmref, r = jvmref, dups(fis .vals .nonnullrefs), hasSharing(fis .vals .nonnullrefs, st), junkref(str, r, addobj(r, class, fis, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 85:
(addobj(r, class, fis, st)[(addobj(r, class, fis, st)[r -- sks .first]) .refval -- sks .rest]) = refval(r0), fis .keys = instfields(class), (addobj(r, class, fis, st)[r -- sks0]) = refval(r0), validfieldinits(str, fis, st), newref(r, st), fits(r, sks0, addobj(r, class, fis, st)), fits(r, sks, addobj(r, class, fis, st)), disjoint(str, fis .vals .nonnullrefs, st), is_referencevalue(addobj(r, class, fis, st)[r -- sks]) ⊦ sks = [], r = jvmref, r0 = jvmref, sks0 = [], sks0 = sks, junkref(str, r, addobj(r, class, fis, st)), hasSharing(fis .vals .nonnullrefs, st), dups(fis .vals .nonnullrefs)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]