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Proof step 10 from proof nonnullrefs-dups (previous step)

sk ≠ sk1, ((sk1 × refval(r1)) ' + fis0)[sk0] = refval(r), fis0[sk] = refval(r), sk0 ∈ (sk1 × refval(r1)) ' + fis0, sk ∈ (sk1 × refval(r1)) ' + fis0, ∀ sk2, r0, sk3. fis0[sk2] = refval(r0) ∧ fis0[sk3] = refval(r0) ∧ sk2 ∈ fis0 ∧ sk3 ∈ fis0 → sk3 = sk2 ∨ r0 = jvmref ⊦ sk = sk0, r = jvmref, r1 = jvmref, dups(fis0 .vals .nonnullrefs), r1 ∈ fis0 .vals .nonnullrefs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
((sk1 × refval(r1)) ' + fis0)[sk0] = refval(r), fis0[sk] = refval(r), sk0 ∈ (sk1 × refval(r1)) ' + fis0, sk ∈ fis0, ∀ sk2, r0, sk3. fis0[sk2] = refval(r0) ∧ fis0[sk3] = refval(r0) ∧ sk2 ∈ fis0 ∧ sk3 ∈ fis0 → sk3 = sk2 ∨ r0 = jvmref ⊦ sk = sk0, r = jvmref, r1 = jvmref, sk = sk1, dups(fis0 .vals .nonnullrefs), r1 ∈ fis0 .vals .nonnullrefs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, fieldinit-assoclista, a, ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPNAMEDSEQ : assoclist, fieldinit, inkeys-n, ⊦ sk ≠ sk0 → ( sk ∈ (sk0 × val) ' + fis ↔ sk ∈ fis)])]]