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Proof step 21 from proof pathex-def-simple (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
⊦ path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, st[refs .first -- _type] .type, st))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]