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Proof step 11 from proof prefixn-append-slice (conclusion)

⊦ n ≤ n0 → prefixn(n, x) + slice(n, n0, x) = prefixn(n0, 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(slice,,, ⊦ slice(n0, n1, x) = (n0 ≤ n1 ⊃ prefixn(n1 - n0, dropn(n0, x)) ;[] )) Csimpnamedseq(dropn,,, ⊦ dropn(n, x) = (n ≤ # x ⊃ restn(n, x);[])) Csimpnamedseq(nat,,ls-03, n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0) Csimpnamedseq(nat,,le-ls-simp, ⊦ m ≤ n → (n < m ↔ false)) Csimpnamedseq(nat,,lels-01, ⊦ ¬ m ≤ n ↔ n < m)))]