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Proof step 43 from proof nonnullrefs-dups-key (previous step)

dups(fis .vals .nonnullrefs) ⊦ dups(fis .keys), ∃ r, sk, sk0. sk ≠ sk0 ∧ r ≠ jvmref ∧ fis[sk] = refval(r) ∧ fis[sk0] = refval(r) ∧ sk ∈ fis ∧ sk0 ∈ fis

User applied rule structural induction

And obtained the following new goals

Proof step 2:
⊦ [] .vals = []
Proof step 42:
dups(fis0 .vals .nonnullrefs) → dups(fis0 .keys) ∨ (∃ r, sk, sk0. sk ≠ sk0 ∧ r ≠ jvmref ∧ fis0[sk] = refval(r) ∧ fis0[sk0] = refval(r) ∧ sk ∈ fis0 ∧ sk0 ∈ fis0), dups((fv + fis0) .vals .nonnullrefs) ⊦ dups((fv + fis0) .keys), ∃ r, sk, sk0. sk ≠ sk0 ∧ r ≠ jvmref ∧ (fv + fis0)[sk] = refval(r) ∧ (fv + fis0)[sk0] = refval(r) ∧ sk ∈ fv + fis0 ∧ sk0 ∈ fv + fis0

Rule argument

[VARWITHVARSEQSARG : fis, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : fieldinits generated by [], +;], [CSIMPNAMEDSEQ : valuesfuns, , nonnullrefs-empty, ⊦ [] .nonnullrefs = []], [CSIMPNAMEDSEQ : list-dup, references, e-01, ⊦ ¬ dups([])])]]