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Proof step 4 from proof firstn-at-in-dups (previous step)

x[n] ∈ x, n < # x, x[n] ∈ firstn(n, x) ⊦ x = [], ∃ m, n. m < # x ∧ n < # x ∧ m ≠ n ∧ x[m] = x[n]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ a ∈ x ↔ (∃ n. n < # x ∧ x[n] = a)
Proof step 6:
x[n] ∈ x, n < # x, ∃ n0. n0 < # firstn(n, x) ∧ firstn(n, x)[n0] = x[n] ⊦ x = [], ∃ m, n. m < # x ∧ n < # x ∧ m ≠ n ∧ x[m] = x[n]

Rule argument

spec: inst: name: in-get seq: ⊦ a ∈ x ↔ (∃ n. n < # x ∧ x[n] = a) subst: [x, a] → [firstn(n, x), x[n]] flags: rewrite, current

Additional information

Used terms: firstn(n, x), x[n]