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

dups(fv .2 ' .nonnullrefs + fis0 .vals .nonnullrefs), dups(fis0 .vals .nonnullrefs) → sk ≠ sk0 ∧ r ≠ jvmref ∧ fis0[sk] = refval(r) ∧ fis0[sk0] = refval(r) ∧ sk ∈ fis0 ∧ sk0 ∈ fis0 ⊦ dups(fis0 .keys), fv .1 ∈ fis0 .keys, ∃ r0, sk1, sk2. sk1 ≠ sk2 ∧ r0 ≠ jvmref ∧ (fv ' + fis0)[sk1] = refval(r0) ∧ (fv ' + fis0)[sk2] = refval(r0) ∧ (sk1 ∈ fv ' ∨ sk1 ∈ fis0) ∧ (sk2 ∈ fv ' ∨ sk2 ∈ fis0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

[x0lemmaarg: spec-name: pair instance-name: fieldinit lemma-name: elim antecedent: succedent: sk = fv .1 ∧ val = fv .2 ↔ fv = sk × val substlist: [fv, sk, val] → [fv, sk3, val] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ sk = fv .1 ∧ val = fv .2 ↔ fv = sk × val])], [EXTRATERMS : (fv, sk3, val)]]