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Proof step 15 from proof nonnullrefs-in (previous step)

∀ sk, r. fis0[sk] = refval(r) ∧ sk ∈ fis0 → r = jvmref ∨ r ∈ fis0 .vals .nonnullrefs, (fv + fis0)[sk] = refval(r), sk ∈ fv + fis0 ⊦ r = jvmref, r ∈ (fv + fis0) .vals .nonnullrefs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ (fv ' + fis) .vals = fv .2 ' + fis .vals
Proof step 14:
(fv ' + fis0)[sk] = refval(r), sk ∈ fv ' + fis0, ∀ sk, r. fis0[sk] = refval(r) ∧ sk ∈ fis0 → r = jvmref ∨ r ∈ fis0 .vals .nonnullrefs ⊦ r = jvmref, r ∈ (fv .2 ' + fis0 .vals) .nonnullrefs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, fieldinit-assoclista, a, ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPNAMEDSEQ : list, fieldinit-assoclista, c, ⊦ fv + fis = fv ' + fis])]]