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Proof step 6 from proof validrefs-nodups-01 (previous step)

¬ ¬ validrefs(str, refs2 + r ' + refs0, ty, st), validrefs(str, refs2 + r ' + refs1 + r ' + refs0, ty, st), ∀ st, ty, str, refs3. # refs3 < # (refs2 + (r + (refs1 + (r + refs0)))) → ( validrefs(str, refs3, ty, st) ∧ refs3 ≠ [] → (∃ refs1. validrefs(str, refs1, ty, st) ∧ refs1 ≠ [] ∧ refs1 .last = refs3 .last ∧ refs3 .first = refs1 .first ∧ ¬ dups(refs1))) ⊦ ∃ refs. validrefs(str, refs, ty, st) ∧ refs ≠ [] ∧ refs .last = (r ' + refs0) .last ∧ (refs2 + r ') .first = refs .first ∧ ¬ dups(refs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
validrefs(str, refs2 + r ' + refs1 + r ' + refs0, ty, st), validrefs(str, refs2 + r ' + refs0, ty, st), ∀ st, ty, str, refs3. # refs3 < # (refs2 + (r + (refs1 + (r + refs0)))) → ( validrefs(str, refs3, ty, st) ∧ refs3 ≠ [] → (∃ refs1. validrefs(str, refs1, ty, st) ∧ refs1 ≠ [] ∧ refs1 .last = refs3 .last ∧ refs3 .first = refs1 .first ∧ ¬ dups(refs1))) ⊦ ∃ refs. validrefs(str, refs, ty, st) ∧ refs ≠ [] ∧ refs .last = (r ' + refs0) .last ∧ (refs2 + r ') .first = refs .first ∧ ¬ dups(refs)

Rule argument

[emptyarg]

Additional information

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