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Proof step 4 from proof pathex-def-nodups (previous step)

∃ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
refs .last = refs1 .last, validrefs(str, refs .first ' + refs1, st[refs .first -- _type] .type, st), dups(refs1), validrefs(str, refs, st[refs .first -- _type] .type, st) ⊦ refs = [], refs .first = refs1 .last, refs1 = [], dups(refs), ∃ refs0. refs0 ≠ [] ∧ refs0 .first = refs .first ∧ refs0 .last = refs1 .last ∧ ¬ dups(refs0) ∧ validrefs(str, refs0, st[refs0 .first -- _type] .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (r ' + refs) .first = r], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ refs ≠ [] → (r ' + refs) .last = refs .last])]]