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Proof step 9 from proof dropn-first (conclusion)

⊦ n ≠ 0 → dropn(n, x) = (x = [] ⊃ [];dropn(n - 1, x.rest))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ dropn(n, x) = (n ≤ # x ⊃ restn(n, x);[])

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(dropn,,, ⊦ dropn(n, x) = (n ≤ # x ⊃ restn(n, x);[])) Csimpnamedseq(nat,,ls-03, n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0) Csimpnamedseq(nat-basic1,,one-01, ⊦ m < 1 ↔ m = 0) Csimpnamedseq(nat,,lels-01, ⊦ ¬ m ≤ n ↔ n < m) Csimpnamedseq(nat,,pred-03, ⊦ n -1 = n - 1) Csimpnamedseq(list-data,,length-rest, ⊦ x ≠ [] → # x.rest = (# x)-1)))]