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Proof step 40 from proof nonnullrefs-dups-key (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
dups(val ' .nonnullrefs + fis0 .vals .nonnullrefs), dups(fis0 .vals .nonnullrefs) → sk ≠ sk0 ∧ r ≠ jvmref ∧ fis0[sk] = refval(r) ∧ fis0[sk0] = refval(r) ∧ sk ∈ fis0 ∧ sk0 ∈ fis0 ⊦ dups(fis0 .keys), sk3 ∈ fis0 .keys, ∃ r0, sk1, sk2. sk1 ≠ sk2 ∧ r0 ≠ jvmref ∧ ((sk3 × val) ' + fis0)[sk1] = refval(r0) ∧ ((sk3 × val) ' + fis0)[sk2] = refval(r0) ∧ (sk1 = sk3 ∨ sk1 ∈ fis0) ∧ (sk2 = sk3 ∨ sk2 ∈ fis0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : assoclist, fieldinit, inone, ⊦ sk ∈ (sk0 × val) ' ↔ sk = sk0])]]