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Proof step 6 from proof nonnullrefs-append (previous step)

is_referencevalue(val), ∀ vals0. (vals1 + vals0) .nonnullrefs = vals1 .nonnullrefs + vals0 .nonnullrefs ⊦ val .refval = jvmref, (vals1 + vals0) .nonnullrefs = vals1 .nonnullrefs + vals0 .nonnullrefs

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
val = refval(r), is_referencevalue(val), ∀ vals0. (vals1 + vals0) .nonnullrefs = vals1 .nonnullrefs + vals0 .nonnullrefs ⊦ r = jvmref, (vals1 + vals0) .nonnullrefs = vals1 .nonnullrefs + 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, r] 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, r)]]