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Proof step 52 from proof hasSharing-addobj-same (previous step)

fis[sk] = refval(r1) ∧ sk ∈ fis ∧ r1 ≠ jvmref, fis[sk0] = refval(r2), fis[sk] = refval(r1), st[r1 -- sks1] = refval(r0), fis .keys = instfields(class), st[r2 -- sks] = refval(r0), fits(r1, sks1, st), sk ∈ instfields(class), is_referencevalue(addobj(r, class, fis, st)[r -- (sk ' + sks1)]), is_referencevalue(addobj(r, class, fis, st)[r -- (sk0 ' + sks)]), sk ∈ fis, validfieldinits(str, fis, st), newref(r, st), fits(r2, sks, st), sk0 ∈ instfields(class), sk0 ∈ fis, r1 -- sks1 ∈ st, r2 -- sks ∈ st, validreflist(str, mkjavavalues(fis .vals .nonnullrefs), mkclasstype(`java.lang.Object' .class), st), ¬ true ∨ ¬ r2 ∈ fis .vals .nonnullrefs ⊦ r0 = jvmref, sk0 ' + sks = sk ' + sks1, r = jvmref, sk = sk0, r1 = r2, junkref(str, r2, st), junkref(str, r1, st), junkref(str, r, addobj(r, class, fis, st)), hasSharing(fis .vals .nonnullrefs, st), dups(fis .vals .nonnullrefs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 48:
⊦ ¬ junkref(str, r, st) ∧ fits(r, sks, st) ∧ r ≠ jvmref → r -- sks ∈ st
Proof step 51:
fis[sk0] = refval(r2), fis[sk] = refval(r1), st[r1 -- sks1] = refval(r0), fis .keys = instfields(class), st[r2 -- sks] = refval(r0), fits(r1, sks1, st), sk ∈ instfields(class), is_referencevalue(addobj(r, class, fis, st)[r -- (sk ' + sks1)]), is_referencevalue(addobj(r, class, fis, st)[r -- (sk0 ' + sks)]), sk ∈ fis, validfieldinits(str, fis, st), newref(r, st), fits(r2, sks, st), sk0 ∈ instfields(class), sk0 ∈ fis, r2 -- sks ∈ st, validreflist(str, mkjavavalues(fis .vals .nonnullrefs), mkclasstype(`java.lang.Object' .class), st) ⊦ r0 = jvmref, sk0 ' + sks = sk ' + sks1, r = jvmref, sk = sk0, r1 = r2, r1 = jvmref, junkref(str, r2, st), junkref(str, r1, st), junkref(str, r, addobj(r, class, fis, st)), hasSharing(fis .vals .nonnullrefs, st), dups(fis .vals .nonnullrefs), r2 ∈ fis .vals .nonnullrefs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]