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Proof step 37 from proof nonnullrefs-dups-firstoc (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
val ' + vals0 = vals2 + refval(r) ' + vals1 + refval(r) ' + vals, ¬ is_referencevalue(val) ∨ val .refval = jvmref ⊦ r = jvmref, refval(r) ∈ vals2, 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, ¬ val ∈ vals2 ∧ val ≠ refval(r) ∧ ¬ val ∈ vals1 ∧ ¬ val ∈ vals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, javavalues, in-01, ⊦ val ≠ val0 → ( val ∈ val0 ' + vals ↔ val ∈ vals)], [CSIMPNAMEDSEQ : list, javavalues, in-17, ⊦ val ∈ val0 ' ↔ val = val0], [CSIMPNAMEDSEQ : list, javavalues, in-append, ⊦ val ∈ vals + vals0 ↔ ¬ ¬ ( val ∈ vals ∨ val ∈ vals0)], [CSIMPNAMEDFORWARD : list, javavalues, in-14, ⊦ val ' + vals = vals0 → val ∈ vals0], [CSIMPNAMEDSEQ : list, javavalues, a, ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1])]]