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Proof step 1 from proof prefixn-size-prefixn (conclusion)

⊦ prefixn(# prefixn(n, x), x) = prefixn(n, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x)
Proof step 3:
⊦ (# (n ≤ # x ⊃ firstn(n, x);x) ≤ # x ⊃ firstn(# (n ≤ # x ⊃ firstn(n, x);x), x);x) = (n ≤ # x ⊃ firstn(n, x);x)

Rule argument

Additional information

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