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Proof step 58 from proof maxpath-object (previous step)

maxpath(str, r, st) = 2 + # refs, st[r -- fs '] = refval(refs0 .first), maxpath(str, refs0 .first, st) = # refs0, # refs + 1 < # refs0, is_obj(str, r, st), fs ' ∈ instfields(st[r -- _type] .type .class), validrefs(str, r ' + refs + r1 ', st[r -- _type] .type, st), maxpath(str, r, st) ≤ # refs0, validrefs(str, refs0, st[refs0 .first -- _type] .type, st), is_valid_ref(str, refs0 .first, st) ⊦ refs0 = [], # validrefset(str, refs0 .first, st) = 0, validrefset(str, refs0 .first, st) = ∅, r = r1, # validrefset(str, r, st) = 0, validrefset(str, r, st) = ∅, has_reffield(refs0 .last, st), dups(refs0), dups(refs), r1 ∈ refs, r ∈ refs, has_reffield(r1, st), cyclic(str, r, st), junkref(str, r, 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] → [refs0, 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 : (refs0, r0, refs1)]]