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Proof step 105 from proof hasSharing-addobj-same (previous step)

¬ dups(fis .keys), fis .keys = instfields(class), ∃ r, sk, sk0. sk ≠ sk0 ∧ r ≠ jvmref ∧ fis[sk] = refval(r) ∧ fis[sk0] = refval(r) ∧ sk ∈ fis ∧ sk0 ∈ fis, newref(r, st), validfieldinits(str, fis, st) ⊦ r = jvmref, ∃ sks1, sks2. fits(r, sks1, addobj(r, class, fis, st)) ∧ fits(r, sks2, addobj(r, class, fis, st)) ∧ sks1 ≠ sks2 ∧ (addobj(r, class, fis, st)[r -- sks1]) .refval ≠ jvmref ∧ (addobj(r, class, fis, st)[r -- sks1]) = (addobj(r, class, fis, st)[r -- sks2]), junkref(str, r, addobj(r, class, fis, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 104:
fis .keys = instfields(class), fis[sk0] = refval(r0), fis[sk] = refval(r0), sk0 ∈ fis, sk ∈ fis, newref(r, st), validfieldinits(str, fis, st) ⊦ r = jvmref, r0 = jvmref, sk0 = sk, junkref(str, r, addobj(r, class, fis, st)), dups(instfields(class)), ∃ sks1, sks2. fits(r, sks1, addobj(r, class, fis, st)) ∧ fits(r, sks2, addobj(r, class, fis, st)) ∧ sks1 ≠ sks2 ∧ (addobj(r, class, fis, st)[r -- sks1]) .refval ≠ jvmref ∧ (addobj(r, class, fis, st)[r -- sks1]) = (addobj(r, class, fis, st)[r -- sks2])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]