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Proof step 26 from proof nonnullrefs-dups-key (previous step)

is_referencevalue(val), dups(fis0 .vals .nonnullrefs) → sk ≠ sk0 ∧ r ≠ jvmref ∧ fis0[sk] = refval(r) ∧ fis0[sk0] = refval(r) ∧ sk ∈ fis0 ∧ sk0 ∈ fis0 ⊦ val .refval = jvmref, dups(fis0 .keys), sk3 ∈ fis0 .keys, ∃ r0, sk1, sk2. sk1 ≠ sk2 ∧ r0 ≠ jvmref ∧ ((sk3 × val) ' + fis0)[sk1] = refval(r0) ∧ ((sk3 × val) ' + fis0)[sk2] = refval(r0) ∧ (sk1 = sk3 ∨ sk1 ∈ fis0) ∧ (sk2 = sk3 ∨ sk2 ∈ fis0), ¬ val .refval ∈ fis0 .vals .nonnullrefs ∧ ¬ dups(fis0 .vals .nonnullrefs)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 25:
val = refval(r1), is_referencevalue(val), dups(fis0 .vals .nonnullrefs) → sk ≠ sk0 ∧ r ≠ jvmref ∧ fis0[sk] = refval(r) ∧ fis0[sk0] = refval(r) ∧ sk ∈ fis0 ∧ sk0 ∈ fis0 ⊦ r1 = jvmref, dups(fis0 .keys), sk3 ∈ fis0 .keys, ∃ r0, sk1, sk2. sk1 ≠ sk2 ∧ r0 ≠ jvmref ∧ ((sk3 × val) ' + fis0)[sk1] = refval(r0) ∧ ((sk3 × val) ' + fis0)[sk2] = refval(r0) ∧ (sk1 = sk3 ∨ sk1 ∈ fis0) ∧ (sk2 = sk3 ∨ sk2 ∈ fis0), ¬ r1 ∈ fis0 .vals .nonnullrefs ∧ ¬ dups(fis0 .vals .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, r1] 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, r1)]]