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Proof step 28 from proof pathex-def-nodups (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
⊦ r = r0, path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ ¬ dups(refs) ∧ validrefs(str, refs, st[refs .first -- _type] .type, st))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]