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Proof step 5 from proof pathex-def-nodups (previous step)

dups(refs1), validrefs(str, r ' + refs1, st[r -- _type] .type, st) ⊦ refs1 = [], r = refs1 .last, ∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = refs1 .last ∧ ¬ dups(refs) ∧ validrefs(str, refs, st[refs .first -- _type] .type, st)

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
validrefs(str, refs, ty, st) ∧ refs ≠ [] ⊦ ∃ refs0. validrefs(str, refs0, ty, st) ∧ refs0 ≠ [] ∧ refs0 .last = refs .last ∧ refs .first = refs0 .first ∧ ¬ dups(refs0)
Proof step 4:
∃ refs0. validrefs(str, refs0, st[r -- _type] .type, st) ∧ refs0 ≠ [] ∧ refs0 .last = (r ' + refs1) .last ∧ (r ' + refs1) .first = refs0 .first ∧ ¬ dups(refs0), dups(refs1), validrefs(str, r ' + refs1, st[r -- _type] .type, st) ⊦ refs1 = [], r = refs1 .last, ∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = refs1 .last ∧ ¬ dups(refs) ∧ validrefs(str, refs, st[refs .first -- _type] .type, st)

Rule argument

[LEMMAARG : validrefs-nodups-01, [str, refs, ty, st] → [str, r ' + refs1, st[r -- _type] .type, st]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ refs + refs0 ≠ [] ↔ refs = [] ∧ refs0 = []], [CSIMPSEQ : ⊦ r ' ≠ []], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1])], [EXTRATERMS : (str, r ' + refs1, st[r -- _type] .type, st)]]