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Proof step 46 from proof fits-no-refval (previous step)

sks0 ⊑ sks .butlast, fits(r, sks, st) ⊦ sks = [], is_referencevalue(st[r -- sks0]), junkref(str, r, st)

User applied rule insert elim lemma

And obtained the following new goals

Proof step 45:
sks = sks1 + sk ', sks0 ⊑ sks1, fits(r, sks, st) ⊦ sks = [], is_referencevalue(st[r -- sks0]), junkref(str, r, st)

Rule argument

[x0lemmaarg: spec-name: list-last instance-name: storekeys lemma-name: last-elim antecedent: sks ≠ [] succedent: sk = sks .last ∧ sks0 = sks .butlast ↔ sks = sks0 + sk ' substlist: [sks, sk, sks0] → [sks, sk, sks1] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: F replace all?: F paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : sks ≠ [] ⊦ sk = sks .last ∧ sks0 = sks .butlast ↔ sks = sks0 + sk '])], [EXTRATERMS : (sks, sk, sks1)]]