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Proof step 18 from proof nodups-butlastn (previous step)

a ∈ butlastn(n, a0 + x), n ≤ #(a0 + x), ∀ a, n. a ∈ butlastn(n, x) ∧ n ≤ # x → a ∈ x ⊦ a ∈ a0 + x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
a ∈ butlastn(n, a0 ' + x), n ≤ 1 + # x, ∀ a, n. a ∈ butlastn(n, x) ∧ n ≤ # x → a ∈ x ⊦ a = a0, a ∈ x

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' + x ↔ ¬(a ≠ b ∧ ¬ a ∈ x); (ss)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)