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Proof step 62 from proof validrefs-subset-01 (previous step)

acc .first = st[r0 -- i '] .refval, st0[r0 -- _type] .type ≤ ty, is_array(str, r0, st), st ⊆ st0, st[r0 -- _type] .type ≤ ty, i < st[r0 -- _length] .intval, is_referencevalue(st[r0 -- i ']), validrefs(str, acc, st[r0 -- _type] .type .type, st), 0 ≤ i, 0 < st[r0 -- _length] .intval, ∀ str0, ty0, st1, st2. validrefs(str0, acc, ty0, st1) ∧ st1 ⊆ st2 ∧ ¬ junkref(str0, acc .first, st1) → validrefs(str0, acc, ty0, st2) ⊦ r0 = jvmref, acc = [], junkref(str, r0, st), ∃ i0. ¬ i0 < 0 ∧ ¬ st0[r0 -- _length] .intval ≤ i0 ∧ is_referencevalue(st0[r0 -- i0 ']) ∧ acc .first = st0[r0 -- i0 '] .refval ∧ validrefs(str, acc, st0[r0 -- _type] .type .type, st0)

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] → [acc, r, refs] ,T ,list-data ,references-olista ,elim-list-c ]

Additional information

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