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Proof step 138 from proof infinite-append-prefixnextli (previous step)

∀ nextli0, lf, n0, ii, x, n, nextli. nextli0(lf(n0)) = [] ∧ ii(0) = 0 ∧ lf(0) = [] ∧ # x = ii(n) ∧ n0 < m0 ∧ (∀ m, n. m ≤ n → ii(m) ≤ ii(n)) ∧ (∀ n. lf(n + 1) = nextli(lf(n))) ∧ (∀ x. nextli(x) = x + nextli0(x)) ∧ (∀ m. n ≤ m → nextli0(lf(m)) = [] ∧ ii(m) = ii(n)) ∧ (∀ n0. lf(n0) = x to ii(n0)) ∧ (∀ n0. ii(n0 + 1) = ii(n0) + # nextli0(lf(n0)) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (x from ii(n0)) to # nextli0(lf(n0)) = nextli0(lf(n0))) → lf(m0) = lf(n0) ∧ nextli0(lf(m0)) = [] ∧ ii(m0) = ii(n0), nextli0(lf(n0)) = [], ii(0) = 0, lf(0) = [], # x = ii(n), n0 < m0 +1, ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextli(lf(n)), ∀ x. nextli(x) = x + nextli0(x), ∀ m. n ≤ m → nextli0(lf(m)) = [] ∧ ii(m) = ii(n), ∀ n0. lf(n0) = x to ii(n0), ∀ n0. ii(n0 + 1) = ii(n0) + # nextli0(lf(n0)) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (x from ii(n0)) to # nextli0(lf(n0)) = nextli0(lf(n0)) ⊦ lf(m0 +1) = lf(n0) ∧ nextli0(lf(m0 +1)) = [] ∧ ii(m0 +1) = ii(n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 139:
nextli0(lf(n0)) = [], ii(0) = 0, lf(0) = [], # x = ii(n), n0 ≤ m0, ∀ nextli0, lf, n0, ii, x, n, nextli. nextli0(lf(n0)) = [] ∧ ii(0) = 0 ∧ lf(0) = [] ∧ # x = ii(n) ∧ n0 < m0 ∧ (∀ m, n. m ≤ n → ii(m) ≤ ii(n)) ∧ (∀ n. lf(n + 1) = nextli(lf(n))) ∧ (∀ x. nextli(x) = x + nextli0(x)) ∧ (∀ m. n ≤ m → nextli0(lf(m)) = [] ∧ ii(m) = ii(n)) ∧ (∀ n0. lf(n0) = x to ii(n0)) ∧ (∀ n0. ii(n0 + 1) = ii(n0) + # nextli0(lf(n0)) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (x from ii(n0)) to # nextli0(lf(n0)) = nextli0(lf(n0))) → lf(m0) = lf(n0) ∧ nextli0(lf(m0)) = [] ∧ ii(m0) = ii(n0), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextli(lf(n)), ∀ x. nextli(x) = x + nextli0(x), ∀ m. n ≤ m → nextli0(lf(m)) = [] ∧ ii(m) = ii(n), ∀ n0. lf(n0) = x to ii(n0), ∀ n0. ii(n0 + 1) = ii(n0) + # nextli0(lf(n0)) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (x from ii(n0)) to # nextli0(lf(n0)) = nextli0(lf(n0)) ⊦ lf(m0 + 1) = lf(n0) ∧ nextli0(lf(m0 + 1)) = [] ∧ ii(m0 + 1) = ii(n0)

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2