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Proof step 8 from proof nonnullrefs-in (previous step)

¬ (¬ is_referencevalue(val) ∨ val .refval = jvmref), fis0[sk] = refval(r), sk ∈ fis0, ∀ sk, r. fis0[sk] = refval(r) ∧ sk ∈ fis0 → r = jvmref ∨ r ∈ fis0 .vals .nonnullrefs ⊦ r = jvmref, sk = sk0, r ∈ (val ' + fis0 .vals) .nonnullrefs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
fis0[sk] = refval(r), is_referencevalue(val), sk ∈ fis0, ∀ sk, r. fis0[sk] = refval(r) ∧ sk ∈ fis0 → r = jvmref ∨ r ∈ fis0 .vals .nonnullrefs ⊦ r = jvmref, sk = sk0, val .refval = jvmref, r = val .refval, r ∈ fis0 .vals .nonnullrefs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, references, a, ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPNAMEDSEQ : list, references, in-06, ⊦ ¬ r ∈ r0 ' + refs ↔ r ≠ r0 ∧ ¬ r ∈ refs], [CSIMPNAMEDSEQ : list, javavalues, a, ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPNAMEDSEQ : list, references, c, ⊦ r + refs = r ' + refs], [CSIMPNAMEDSEQ : valuesfuns, , nonnullrefs-yes, ⊦ is_referencevalue(val) ∧ val .refval ≠ jvmref → (val ' + vals) .nonnullrefs = val .refval + vals .nonnullrefs])]]