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Proof step 2 from proof prefixn-nodups (previous step)

dups(prefixn(n, x)) ⊦ dups(x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x)
Proof step 4:
dups((n ≤ # x ⊃ firstn(n, x);x)) ⊦ dups(x)

Rule argument

Additional information

Used simplifier rules: prefixn: ⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x);