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Proof step 24 from proof nonnullrefs-dups-firstoc (previous step)

dups(vals0 .nonnullrefs) ↔ (∃ r, vals, vals1, vals2. vals0 = vals + refval(r) ' + vals1 + refval(r) ' + vals2 ∧ r ≠ jvmref ∧ ¬ refval(r) ∈ vals ∧ ¬ refval(r) ∈ vals1) ⊦ dups(vals0 .nonnullrefs), r0 = jvmref, r0 ∈ vals0 .nonnullrefs ∨ dups(vals0 .nonnullrefs) ↔ (∃ r, vals, vals1, vals2. refval(r0) ' + 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 23:
⊦ r0 = jvmref, r0 ∈ vals0 .nonnullrefs ↔ (∃ r, vals, vals1, vals2. refval(r0) ' + vals0 = vals + refval(r) ' + vals1 + refval(r) ' + vals2 ∧ r ≠ jvmref ∧ ¬ refval(r) ∈ vals ∧ ¬ refval(r) ∈ vals1), dups(vals0 .nonnullrefs), ∃ r, vals, vals1, vals2. vals0 = vals + refval(r) ' + vals1 + refval(r) ' + vals2 ∧ r ≠ jvmref ∧ ¬ refval(r) ∈ vals ∧ ¬ refval(r) ∈ vals1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]