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Proof step 12 from proof unchanged-cyclic (previous step)

dups(refs), validrefs(str, refs, st[refs .first -- _type] .type, st), unchanged(str, refs .first, st, st0) ⊦ junkref(str, refs .first, st), ∃ refs0. dups(refs0) ∧ refs0 .first = refs .first ∧ validrefs(str, refs0, st0[refs0 .first -- _type] .type, st0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
refs = r + refs1 ∧ refs ≠ [], dups(refs), validrefs(str, refs, st[r -- _type] .type, st), unchanged(str, r, st, st0) ⊦ junkref(str, r, st), ∃ refs0. dups(refs0) ∧ refs0 .first = r ∧ validrefs(str, refs0, st0[refs0 .first -- _type] .type, st0)

Rule argument

[x0lemmaarg: spec-name: list-data instance-name: references lemma-name: elim-list-c antecedent: refs ≠ [] succedent: r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 substlist: [refs, r, refs0] → [refs, r, refs1] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list-dup, references, e-01, ⊦ ¬ dups([])], [CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (refs, r, refs1)]]