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Proof step 41 from proof nonnullrefs-dups (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 40:
val = refval(r0), dups(vals0 .nonnullrefs) ↔ (∃ r, m, n. vals0[m] = refval(r) ∧ vals0[n] = refval(r) ∧ r ≠ jvmref ∧ m ≠ n ∧ m < # vals0 ∧ n < # vals0), is_referencevalue(val) ⊦ r0 = jvmref, r0 ∈ vals0 .nonnullrefs ∨ dups(vals0 .nonnullrefs) ↔ (∃ r, m, n. (val ' + vals0)[m] = refval(r) ∧ (val ' + vals0)[n] = refval(r) ∧ r ≠ jvmref ∧ m ≠ n ∧ ¬ # vals0 < m ∧ ¬ # vals0 < n)

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)]]