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Proof step 10 from proof validrefs-nodups-01 (previous step)

validrefs(str, refs, ty, st), ∀ st, ty, str, refs0. # refs0 < # refs → ( validrefs(str, refs0, ty, st) ∧ refs0 ≠ [] → (∃ refs1. validrefs(str, refs1, ty, st) ∧ refs1 ≠ [] ∧ refs1 .last = refs0 .last ∧ refs0 .first = refs1 .first ∧ ¬ dups(refs1))) ⊦ refs = [], validrefs(str, refs, ty, st) ∧ refs ≠ [] ∧ refs .last = refs .last ∧ refs .first = refs .first ∧ ¬ dups(refs), ∃ refs0. validrefs(str, refs0, ty, st) ∧ refs0 ≠ [] ∧ refs0 .last = refs .last ∧ refs .first = refs0 .first ∧ ¬ dups(refs0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
validrefs(str, refs, ty, st), dups(refs), ∀ st, ty, str, refs0. # refs0 < # refs → ( validrefs(str, refs0, ty, st) ∧ refs0 ≠ [] → (∃ refs1. validrefs(str, refs1, ty, st) ∧ refs1 ≠ [] ∧ refs1 .last = refs0 .last ∧ refs0 .first = refs1 .first ∧ ¬ dups(refs1))) ⊦ refs = [], ∃ refs0. validrefs(str, refs0, ty, st) ∧ refs0 ≠ [] ∧ refs0 .last = refs .last ∧ refs .first = refs0 .first ∧ ¬ dups(refs0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]