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Proof step 8 from proof dups-def (previous step)

a ∈ x0 ⊦ ∃ m, n. ¬ # x0 < m ∧ ¬ # x0 < n ∧ m ≠ n ∧ (a ' + x0)[m] = (a ' + x0)[n], dups(x0), ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ a ∈ x ↔ (∃ n. n < # x ∧ x[n] = a)
Proof step 10:
∃ n. n < # x0 ∧ x0[n] = a ⊦ ∃ m, n. ¬ # x0 < m ∧ ¬ # x0 < n ∧ m ≠ n ∧ (a ' + x0)[m] = (a ' + x0)[n], dups(x0), ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n]

Rule argument

spec: list-del inst: name: in-get seq: ⊦ a ∈ x ↔ (∃ n. n < # x ∧ x[n] = a) subst: [x, a] → [x0, a] flags: rewrite, current

Additional information

Used terms: x0, a