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Proof step 7 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 case distinction

And obtained the following new goals

Proof step 8:
a ∈ x0 ⊦ dups(x0), ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n], ∃ m, n. ¬ # x0 < m ∧ ¬ # x0 < n ∧ m ≠ n ∧ (a ' + x0)[m] = (a ' + x0)[n]
Proof step 12:
∃ m, n. ¬ # x0 < m ∧ ¬ # x0 < n ∧ m ≠ n ∧ (a ' + x0)[m] = (a ' + x0)[n] ⊦ dups(x0), a ∈ x0, ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n]

Rule argument

pos: right 1

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