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Proof step 89 from proof validrefs-subset-equiv (previous step)

refs .first = st0[r -- i '] .refval, is_array(str, r, st), st[r -- _type] .type ≤ ty, st ⊆ st0, st0[r -- _type] .type ≤ ty, i < st0[r -- _length] .intval, is_referencevalue(st0[r -- i ']), validrefs(str, refs, st0[r -- _type] .type .type, st0), 0 ≤ i, 0 < st0[r -- _length] .intval, ∀ st, st0, ty, r, str, refs0. # refs0 < # refs → (validrefs(str, r ' + refs0, ty, st0) ∧ st ⊆ st0 ∧ validref(str, r, st[r -- _type] .type, st) → validrefs(str, r ' + refs0, ty, st)) ⊦ r = jvmref, refs = [], junkref(str, r, st), ∃ i0. ¬ i0 < 0 ∧ ¬ st[r -- _length] .intval ≤ i0 ∧ is_referencevalue(st[r -- i0 ']) ∧ refs .first = st[r -- i0 '] .refval ∧ validrefs(str, refs, st[r -- _type] .type .type, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [refs, r0, refs1] ,T ,list-data ,references-olista ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (refs, r0, refs1)]]