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Proof step 105 from proof infinite-append-prefixnextli (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 15 terms: m abort: F discard: F computed: T

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: m