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Proof step 10 from proof dups-def (previous step)

∃ 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]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n < # x0 ⊦ x0 = [], dups(x0), ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n], ∃ m, n0. ¬ # x0 < m ∧ ¬ # x0 < n0 ∧ m ≠ n0 ∧ (x0[n] ' + x0)[m] = (x0[n] ' + x0)[n0]

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1