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Proof step 14 from proof prefixn-split-sublist (conclusion)

⊦ n0 + n1 = n ∧ n ≤ # x → prefixn(n, x) = prefixn(n0, x) + sublist(n0, n1, 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)) Csimpnamedseq(nat-basic2,,c, ⊦ m + n = n + m) Csimpnamedseq(nat-basic2,,a, ⊦ (m + n0) + n1 = m + n0 + n1) Csimpnamedseq(nat-basic2,,l-02, ⊦ ¬ m + n < m) Csimpnamedforward(nat,,fle-01, ⊦ m ≤ n ∧ n < n0 → m < n0) Csimpnamedforward(nat-basic1,,f, ⊦ m < n → n ≠ 0) Csimpnamedseq(nat,,lels-01, ⊦ ¬ m ≤ n ↔ n < m)))]