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Proof step 78 from proof hasSharing-dd-help (previous step)

# refs4 = # refs0, # sks0 = 2 + # refs2 + # refs0, # refs1 + # refs3 = # refs2, st[r -- firstn(1 + # refs0, sks0)] .refval = refs3 .first, fits(r, sks0, st), validrefs(str, (r ' + refs4 + refs3) + r1 ' + refs1 + r0 ', st[r -- _type] .type, st), validrefs(str, r ' + refs0 + refs3 .first ' + refs2 + r0 ', st[r -- _type] .type, st) ⊦ r1 = r0, r0 = jvmref, refs4 + refs3 = [], refs3 = [], refs3 .first = r1, hasSharing(r, st), path∃(str, refs3 .first, r1, st), path∃(str, r1, refs3 .first, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 77:
refs3 = r2 + refs, # refs4 = # refs0, # sks0 = 2 + # refs2 + # refs0, # refs1 + # refs3 = # refs2, st[r -- firstn(1 + # refs0, sks0)] .refval = r2, fits(r, sks0, st), validrefs(str, (r ' + refs4 + refs3) + r1 ' + refs1 + r0 ', st[r -- _type] .type, st), validrefs(str, r ' + refs0 + r2 ' + refs2 + r0 ', st[r -- _type] .type, st) ⊦ r1 = r0, r0 = jvmref, refs4 + refs3 = [], refs3 = [], r2 = r1, hasSharing(r, st), path∃(str, r2, r1, st), path∃(str, r1, r2, st)

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [refs3, r2, refs] ,T ,list-data ,references ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (refs3, r2, refs)]]