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Proof step 4 from proof prefix-prefixn-not-shorter (conclusion)

⊦ prefixn(n, x0) ⊑ x1 ∧ n < # x0 → (# x1 < n ↔ false)

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)) Csimpnamedseq(nat-basic2,,succ, ⊦ m +1 = m + 1) Csimpnamedseq(list-data,,l, ⊦ # x = 0 ↔ x = []) Csimpnamedforward(nat-basic1,,trans-1, ⊦ n < n0 ∧ n0 < n1 → n < n1) Csimpnamedforward(nat-basic1,,f, ⊦ m < n → n ≠ 0) Csimpnamedforward(nat-basic1,,lf, ⊦ m < n ∧ n < n0 → m +1 < n0) Csimpnamedseq(nat,,lels-03, ⊦ m < n → m ≤ n)))]