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Proof step 48 from proof infinite-append-prefixnextnli (previous step)

# x = ii(n), ii(0) = 0, ii(n + 1) = ii(n), lf(n + 1) = nextnli(lf(n), n), lf(0) = [], nextnli(x, n) = x + nextnli0(x, n), nextnli0(lf(n), n) = [], ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextnli(lf(n), n), ∀ n0. lf(n0) = x to ii(n0) ⊦ nextnli(x, n) = x

User applied rule all left

And closed the goal!

Rule argument

pos: left 10 terms: n abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ # x = n → x to n = x; (s)% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: n