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Proof step 80 from proof validrefs-subset (previous step)

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