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

val = refval(r1), is_referencevalue(val), ∀ r0. r0 ∈ vals0 .nonnullrefs ↔ r0 ≠ jvmref ∧ (∃ m. vals0[m] = refval(r0) ∧ m < # vals0) ⊦ r1 = jvmref, r = r1 ∨ r ∈ vals0 .nonnullrefs ↔ r ≠ jvmref ∧ (∃ m. (val ' + vals0)[m] = refval(r) ∧ ¬ # vals0 < m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
∀ r0. r0 ∈ vals0 .nonnullrefs ↔ r0 ≠ jvmref ∧ (∃ m. vals0[m] = refval(r0) ∧ m < # vals0) ⊦ r1 = jvmref, r = r1 ∨ r ∈ vals0 .nonnullrefs ↔ r ≠ jvmref ∧ (∃ m. (refval(r1) ' + vals0)[m] = refval(r) ∧ ¬ # vals0 < m)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : javavalue, , test-19, ⊦ is_referencevalue(refval(r))])]]