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

n < #(a + x0), dups(lastn(n, a + x0)), ∀ n. n < # x0 ∧ dups(lastn(n, x0)) → dups(x0) ⊦ dups(a + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
dups(lastn(n, x0)), n ≤ # x0, ∀ n0. ¬ n0 < # x0 ∨ ¬ dups(lastn(n0, x0)) ⊦ dups(x0), a ∈ x0

Rule argument

Additional information

Used simplifier rules: lastN-append-01: ⊦ n ≤ # y → lastn(n, x + y) = lastn(n, y);% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x; (ws)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)