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Proof step 4 from proof lastN-dups (previous step)

dups(lastn(n, x0)), n ≤ # x0, ∀ n0. ¬ n0 < # x0 ∨ ¬ dups(lastn(n0, x0)) ⊦ dups(x0), a ∈ 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;% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s) Used terms: n