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Proof step 1 from proof diffgetn-nodupssize (conclusion)

⊦ ¬ dups(x) ∧ n < # x → #(x \ x[n] ') = # x - 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n < # x ⊦ #(x \ x[n] ') = # x - 1, x = [], dups(x)

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1