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Proof step 2 from proof prefixn-le-size (conclusion)

⊦ # x ≤ # prefixn(n, x) ↔ prefixn(n, x) = x ∧ 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(nat,,ler, ⊦ m ≤ m) Csimpnamedseq(nat-basic1,,ref-1, ⊦ ¬ n < n) Csimpnamedseq(list-del,,first-all-01, ⊦ # x = n → firstn(n, x) = x) Csimpnamedseq(list-del,,first-length, ⊦ # firstn(n, x) = n) Csimpnamedseq(nat-basic1,,ls-02, ⊦ m < n → ¬ n < m) Csimpnamedseq(nat,,lels-01, ⊦ ¬ m ≤ n ↔ n < m) Csimpnamedseq(list-del,,first-all, ⊦ firstn(# x, x) = x) Csimpnamedseq(nat,,anti, ⊦ m ≤ n → (n ≤ m ↔ m = n)) Csimpnamedseq(nat,,ge, ⊦ n ≥ m ↔ m ≤ n)))]