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Proof step 22 from proof pathex-def-nodups (previous step)

∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, st[refs .first -- _type] .type, st) ⊦ ∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ ¬ dups(refs) ∧ validrefs(str, refs, st[refs .first -- _type] .type, st), r = r0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
validrefs(str, refs, st[refs .first -- _type] .type, st) ⊦ refs .first = refs .last, refs = [], ∃ refs0. refs0 ≠ [] ∧ refs0 .first = refs .first ∧ refs0 .last = refs .last ∧ ¬ dups(refs0) ∧ validrefs(str, refs0, st[refs0 .first -- _type] .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]