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Proof step 25 from proof pathex-def-nodups (previous step)

⊦ r = r0, (∃ 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))

User applied rule case distinction

And obtained the following new goals

Proof step 22:
∃ 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
Proof step 24:
∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ ¬ dups(refs) ∧ validrefs(str, refs, st[refs .first -- _type] .type, st) ⊦ ∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, st[refs .first -- _type] .type, st), r = r0

Rule argument

[CASEDARG : pos: right 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]