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Proof step 51 from proof hasSharing-pathex (previous step)

(refs1 + r0 ') .first = st[r -- i '] .refval, is_array(str, r, st), hasSharing(r0, st), i < st[r -- _length] .intval, is_referencevalue(st[r -- i ']), validrefs(str, refs1 + r0 ', st[r -- _type] .type .type, st), 0 ≤ i, 0 < st[r -- _length] .intval, ∀ st, r, str, refs. # refs < # (refs1 + r0 ') → (validrefs(str, r ' + refs, st[r -- _type] .type, st) ∧ hasSharing(refs .last, st) → refs = [] ∨ hasSharing(r, st)) ⊦ r = jvmref, hasSharing(r, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 50:
refs1 + r0 ' = r1 + refs0 ∧ refs1 + r0 ' ≠ [], r1 = st[r -- i '] .refval, is_array(str, r, st), hasSharing(r0, st), i < st[r -- _length] .intval, is_referencevalue(st[r -- i ']), validrefs(str, refs1 + r0 ', st[r -- _type] .type .type, st), 0 ≤ i, 0 < st[r -- _length] .intval, ∀ st, r, str, refs. # refs < # (refs1 + r0 ') → (validrefs(str, r ' + refs, st[r -- _type] .type, st) ∧ hasSharing(refs .last, st) → refs = [] ∨ hasSharing(r, st)) ⊦ r = jvmref, hasSharing(r, st)

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [refs1 + r0 ', r1, refs0] ,T ,list-data ,references-olista ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ refs + refs0 ≠ [] ↔ refs = [] ∧ refs0 = []], [CSIMPSEQ : ⊦ r ' ≠ []], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (refs1 + r0 ', r1, refs0)]]