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Proof step 6 from proof nonnullrefs-null (previous step)

¬ jvmref ∈ vals0 .nonnullrefs, jvmref ∈ (val + vals0) .nonnullrefs ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (vals + vals0) .nonnullrefs = vals .nonnullrefs + vals0 .nonnullrefs
Proof step 5:
jvmref ∈ val ' .nonnullrefs ⊦ jvmref ∈ vals0 .nonnullrefs

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])]]