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Proof step 17 from proof nonnullrefs-in-firstoc (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 16:
val = refval(r0), is_referencevalue(val), r ∈ vals0 .nonnullrefs → r ≠ jvmref ∧ vals0 = vals + refval(r) ' + vals1 ∧ ¬ refval(r) ∈ vals ⊦ r0 = jvmref, r ≠ jvmref ∧ (∃ vals, vals1. val ' + vals0 = vals + refval(r) ' + vals1 ∧ ¬ refval(r) ∈ vals), r ≠ r0 ∧ ¬ r ∈ vals0 .nonnullrefs

Rule argument

[x0lemmaarg: spec-name: javavalue instance-name: lemma-name: elim-13 antecedent: succedent: is_referencevalue(val) → (r = val .refval ↔ val = refval(r)) substlist: [val, r] → [val, r0] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_referencevalue(val) → ( r = val .refval ↔ val = refval(r))])], [EXTRATERMS : (val, r0)]]