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Proof step 22 from proof nonnullrefs-in-firstoc (previous step)

¬ is_referencevalue(val) ∨ val .refval = jvmref, r ∈ vals0 .nonnullrefs → r ≠ jvmref ∧ vals0 = vals + refval(r) ' + vals1 ∧ ¬ refval(r) ∈ vals ⊦ ¬ r ∈ [] ∧ ¬ r ∈ vals0 .nonnullrefs, r ≠ jvmref ∧ (∃ vals, vals1. val ' + vals0 = vals + refval(r) ' + vals1 ∧ ¬ refval(r) ∈ vals)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
⊦ (vals + vals0) .nonnullrefs = vals .nonnullrefs + vals0 .nonnullrefs
Proof step 20:
⊦ r ≠ jvmref → refval(r) ' .nonnullrefs = r '
Proof step 21:
¬ is_referencevalue(val) ∨ val .refval = jvmref ⊦ r = jvmref, refval(r) ∈ vals, ∃ vals3, vals2. val ' + vals + refval(r) ' + vals1 = vals3 + refval(r) ' + vals2 ∧ ¬ refval(r) ∈ vals3

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, references, a, ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPNAMEDSEQ : list, references, in-17, ⊦ r ∈ r0 ' ↔ r = r0], [CSIMPNAMEDSEQ : list, references, in-append, ⊦ r ∈ refs + refs0 ↔ ¬ ¬ (r ∈ refs ∨ r ∈ refs0)], [CSIMPNAMEDSEQ : list, javavalues, a, ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPNAMEDSEQ : list, references, in-empty, ⊦ ¬ r ∈ []])]]