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

n ≤ # x0, dups(lastn(n, x0)), ∀ n0. n0 < # x0 → ¬ dups(lastn(n0, x0)) ⊦ a ∈ x0, dups(x0)

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n abort: F discard: F computed: T

Additional information

Used simplifier rules: lastN-all: ⊦ n = # x → lastn(n, x) = x;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat Used terms: n