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Proof step 35 from proof nonnullrefs-dups-key (previous step)

fis0[sk] = refval(r), fis0[sk0] = refval(r), sk ∈ fis0, sk0 ∈ fis0, ¬ is_referencevalue(val) ∨ val .refval = jvmref ⊦ r = jvmref, sk = sk0, dups(fis0 .keys), sk3 ∈ fis0 .keys, ((sk3 × val) ' + fis0)[sk] = refval(r) ∧ ((sk3 × val) ' + fis0)[sk0] = refval(r)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 29:
⊦ sk ≠ sk0 → ((sk0 × val) ' + fis)[sk] = fis[sk]
Proof step 30:
⊦ ¬ sk ∈ fis .keys → ¬ sk ∈ fis
Proof step 34:
sk ≠ sk3, fis0[sk] = refval(r), fis0[sk0] = refval(r), sk ∈ fis0, sk0 ∈ fis0, ¬ is_referencevalue(val) ∨ val .refval = jvmref ⊦ r = jvmref, sk = sk0, dups(fis0 .keys), sk3 ∈ fis0 .keys, fis0[sk] = refval(r) ∧ ((sk3 × val) ' + fis0)[sk0] = refval(r)

Rule argument

[x0lemmaarg: spec-name: assoclist instance-name: fieldinit lemma-name: assoc-no antecedent: succedent: sk ≠ sk0 → ((sk0 × val) ' + fis)[sk] = fis[sk] substlist: [sk, fis, val, sk0] → [sk, fis0, val, sk3] add preconditions?: F is rewrite lemma?: T current spec?: F rotate lemma?: F replace all?: T paths: ((1, 0, 4, 1, 0)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk, fis0, val, sk3)]]