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Proof step 26 from proof nonnullrefs-in-firstoc (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (vals + vals0) .nonnullrefs = vals .nonnullrefs + vals0 .nonnullrefs
Proof step 25:
∀ r. r ∈ vals0 .nonnullrefs → r ≠ jvmref ∧ (∃ vals2, vals1. vals0 = vals2 + refval(r) ' + vals1 ∧ ¬ refval(r) ∈ vals2) ⊦ ¬ r ∈ val ' .nonnullrefs ∧ ¬ r ∈ vals0 .nonnullrefs, r ≠ jvmref ∧ (∃ vals, vals1. val ' + vals0 = vals + refval(r) ' + vals1 ∧ ¬ refval(r) ∈ vals)

Rule argument

[emptyarg]

Additional information

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