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Proof step 3 from proof firstn-at-in-dups (previous step)

n < # x, x[n] ∈ x, x[n] ∈ firstn(n, x) ⊦ x = [], dups(x)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ dups(x) ↔ (∃ m, n. m < # x ∧ n < # x ∧ m ≠ n ∧ x[m] = x[n])
Proof step 5:
n < # x, x[n] ∈ firstn(n, x), x[n] ∈ x ⊦ x = [], ∃ m, n. m < # x ∧ n < # x ∧ m ≠ n ∧ x[m] = x[n]

Rule argument

spec: <toplevel> name: dups-def seq: ⊦ dups(x) ↔ (∃ m, n. m < # x ∧ n < # x ∧ m ≠ n ∧ x[m] = x[n]) subst: [x] → [x] flags: <all paths>

Additional information

Used terms: x