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Proof step 77 from proof path-dups (previous step)

dups(x), path(x, H) ⊦ cyclic(hd x, H)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ dups(x0) ↔ (∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n])
Proof step 76:
∃ m, n. m < # x ∧ n < # x ∧ m ≠ n ∧ x[m] = x[n], path(x, H) ⊦ cyclic(hd x, H)

Rule argument

[x0lemmaarg: spec-name: list-del instance-name: reflist lemma-name: dups-def antecedent: succedent: dups(x0) ↔ (∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n]) substlist: [x0] → [x] add preconditions?: F is rewrite lemma?: T current spec?: F rotate lemma?: F replace all?: T paths: ((0, 0, 0)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (x)]]