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Proof step 28 from proof nonnullrefs-dups-firstoc (previous step)

¬ (¬ is_referencevalue(val) ∨ val .refval = jvmref), dups(vals0 .nonnullrefs) ↔ (∃ r, vals, vals1, vals2. vals0 = vals + refval(r) ' + vals1 + refval(r) ' + vals2 ∧ r ≠ jvmref ∧ ¬ refval(r) ∈ vals ∧ ¬ refval(r) ∈ vals1) ⊦ dups((val ' + vals0) .nonnullrefs) ↔ (∃ r, vals, vals1, vals2. val ' + vals0 = vals + refval(r) ' + vals1 + refval(r) ' + vals2 ∧ r ≠ jvmref ∧ ¬ refval(r) ∈ vals ∧ ¬ refval(r) ∈ vals1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ is_referencevalue(val) ∧ val .refval ≠ jvmref → (val ' + vals) .nonnullrefs = val .refval + vals .nonnullrefs
Proof step 27:
dups(vals0 .nonnullrefs) ↔ (∃ r, vals, vals1, vals2. vals0 = vals + refval(r) ' + vals1 + refval(r) ' + vals2 ∧ r ≠ jvmref ∧ ¬ refval(r) ∈ vals ∧ ¬ refval(r) ∈ vals1), is_referencevalue(val) ⊦ val .refval = jvmref, val .refval ∈ vals0 .nonnullrefs ∨ dups(vals0 .nonnullrefs) ↔ (∃ r, vals, vals1, vals2. val ' + vals0 = vals + refval(r) ' + vals1 + refval(r) ' + vals2 ∧ r ≠ jvmref ∧ ¬ refval(r) ∈ vals ∧ ¬ refval(r) ∈ vals1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, references, a, ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPNAMEDSEQ : list-dup, references, rec, ⊦ dups(r ' + refs) ↔ ¬ ¬ (r ∈ refs ∨ dups(refs))], [CSIMPNAMEDSEQ : list, javavalues, a, ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPNAMEDSEQ : list, references, c, ⊦ r + refs = r ' + refs])]]