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Proof step 2 from proof prefixn-append-left (conclusion)

⊦ n ≤ # x → prefixn(n, x + y) = prefixn(n, 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(list-del,,first-append, ⊦ m ≤ # x → firstn(m, x + y) = firstn(m, 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-basic1,,f, ⊦ m < n → n ≠ 0) Csimpnamedforward(nat,,fle, ⊦ m < n ∧ n ≤ n0 → m < n0) Csimpnamedseq(nat,,lels-01, ⊦ ¬ m ≤ n ↔ n < m) Csimpnamedseq(list,,a, ⊦ (x + y) + z = x + y + z) Csimpnamedseq(list,,len, ⊦ #(x + y) = # x + # y)))]