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Proof step 18 from proof nonnullrefs-in-firstoc (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ is_referencevalue(val) ∧ val .refval ≠ jvmref → val ' .nonnullrefs = val .refval '
Proof step 17:
is_referencevalue(val), r ∈ vals0 .nonnullrefs → r ≠ jvmref ∧ vals0 = vals + refval(r) ' + vals1 ∧ ¬ refval(r) ∈ vals ⊦ val .refval = jvmref, r ≠ jvmref ∧ (∃ vals, vals1. val ' + vals0 = vals + refval(r) ' + vals1 ∧ ¬ refval(r) ∈ vals), r ≠ val .refval ∧ ¬ r ∈ vals0 .nonnullrefs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, references, in-17, ⊦ r ∈ r0 ' ↔ r = r0])]]