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Proof step 44 from proof fits-no-refval (previous step)

sks0 ⊑ sks1, fits(r, sks1 + sk ', st) ⊦ is_referencevalue(st[r -- sks0]), junkref(str, r, st)

User applied rule structural induction

And obtained the following new goals

Proof step 43:
∀ sks0, r, sk, st, str. sks0 ⊑ sks ∧ fits(r, sks + sk ', st) → is_referencevalue(st[r -- sks0]) ∨ junkref(str, r, st), sks0 ⊑ sk0 + sks, fits(r, (sk0 + sks) + sk ', st) ⊦ is_referencevalue(st[r -- sks0]), junkref(str, r, st)

Rule argument

[VARWITHVARSEQSARG : sks1, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : storekeys generated by [], +;], [CSIMPNAMEDSEQ : list-last, storekeys, eq-01-01, ⊦ sks ⊑ [] ↔ sks = []], [CSIMPNAMEDSEQ : list, storekeys, Nil, ⊦ [] + sks = sks], [CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPNAMEDSEQ : storefuns, , lookup-empty, ⊦ st[r -- []] = refval(r)], [CSIMPNAMEDSEQ : javavalue, , test-19, ⊦ is_referencevalue(refval(r))])]]