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Proof step 3 from proof dups-def (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 4:
dups(x0), dups(x0) ↔ (∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n]) ⊦ a ∈ x0 ∨ dups(x0) ↔ (∃ m, n. ¬ # x0 < m ∧ ¬ # x0 < n ∧ m ≠ n ∧ (a ' + x0)[m] = (a ' + x0)[n])
Proof step 6:
dups(x0) ↔ (∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n]) ⊦ dups(x0), a ∈ x0 ∨ dups(x0) ↔ (∃ m, n. ¬ # x0 < m ∧ ¬ # x0 < n ∧ m ≠ n ∧ (a ' + x0)[m] = (a ' + x0)[n])

Rule argument

fma: dups(x0)

Additional information

Used formulas: dups(x0)