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Proof step 15 from proof nonnullrefs-in-key (previous step)

r ∈ fis0 .vals .nonnullrefs → fis0[sk] = refval(r) ∧ sk ∈ fis0 ⊦ r = r0, r = jvmref, r0 = jvmref, dups(fis0 .keys), sk0 ∈ fis0 .keys, ∃ sk. ((sk0 × refval(r0)) ' + fis0)[sk] = refval(r) ∧ (sk = sk0 ∨ sk ∈ fis0), r ≠ r0 ∧ ¬ r ∈ fis0 .vals .nonnullrefs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ fis[sk] = refval(r) ∧ sk ∈ fis ∧ r ≠ jvmref → r ∈ fis .vals .nonnullrefs
Proof step 14:
fis0[sk] = refval(r), sk ∈ fis0 ⊦ r = r0, r = jvmref, r0 = jvmref, dups(fis0 .keys), sk0 ∈ fis0 .keys, ∃ sk. ((sk0 × refval(r0)) ' + fis0)[sk] = refval(r) ∧ (sk = sk0 ∨ sk ∈ fis0)

Rule argument

[emptyarg]

Additional information

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