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Proof step 13 from proof hasSharing-addobj-same (previous step)

¬ sk0 ∈ fis, (addobj(r, class, fis, st)[r2 -- sks]) = refval(r0), fis .keys = instfields(class), (addobj(r, class, fis, st)[r1 -- sks1]) = refval(r0), fis[sk] = refval(r1), (addobj(r, class, fis, st)[r -- sk0]) = refval(r2), sk0 ∈ instfields(class), fits(r2, sks, addobj(r, class, fis, st)), newref(r, st), validfieldinits(str, fis, st), sk ∈ fis, is_referencevalue(addobj(r, class, fis, st)[r -- (sk0 ' + sks)]), disjoint(str, fis .vals .nonnullrefs, st), is_referencevalue(addobj(r, class, fis, st)[r -- (sk ' + sks1)]), sk ∈ instfields(class), fits(r1, sks1, addobj(r, class, fis, st)) ⊦ r = jvmref, sk0 ' + sks = sk ' + sks1, r0 = jvmref, dups(fis .vals .nonnullrefs), hasSharing(fis .vals .nonnullrefs, st), junkref(str, r, addobj(r, class, fis, st))

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : okfuns, , keys-notin, ⊦ ¬ sk ∈ fis ∧ sk ∈ sks → fis .keys ≠ sks])]]