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Proof step 20 from proof prefixn-prefixn (conclusion)

⊦ prefixn(n, prefixn(m, x)) = prefixn((n < m ⊃ n;m), x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x)

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(prefixn,,, ⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x))))]