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Proof step 33 from proof infinite-append-prefixnextli-weak (previous step)

nextli(lf(n)) = lf(n) + nextli0(lf(n)), # lf(n) = ii(n), lf(0) = [], ii(0) = 0, ∀ n0. ii(n0 + 1) = ii(n0) + # nextli0(lf(n0)) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (lf(n) from ii(n0)) to # nextli0(lf(n0)) = nextli0(lf(n0)), ∀ n0. lf(n0) = lf(n) to ii(n0), ∀ m. n ≤ m → nextli0(lf(m)) = [] ∧ ii(m) = ii(n), ∀ x. nextli(x) = x + nextli0(x), ∀ n. lf(n + 1) = nextli(lf(n)), ∀ m, n. m ≤ n → ii(m) ≤ ii(n) ⊦ nextli(lf(n)) = lf(n)

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: n abort: F discard: F computed: T

Additional information

Used simplifier rules: ler: ⊦ m ≤ m; from specification nat%app: ⊦ x + [] = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: n