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Proof step 150 from proof infinite-append-prefixnextli (previous step)

# x = ii(n), ii(0) = 0, ii(m0 + 1) = ii(m0) + # nextli0(lf(m0)), lf(2 + m0) = nextli(lf(m0 + 1)), lf(0) = [], lf(m0 + 1) = lf(m0) + nextli0(lf(m0)), nextli0(lf(n0)) = [], (x from ii(m0)) to # nextli0(lf(m0)) = nextli0(lf(m0)), ii(m0 + 1) ≤ ii(n), n0 ≤ m0, n0 < m0 → lf(m0) = lf(n0) ∧ nextli0(lf(m0)) = [] ∧ ii(m0) = 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)), ∀ n0. lf(n0) = x 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) ⊦ lf(m0 + 1) = lf(n0) ∧ nextli0(lf(m0 + 1)) = [] ∧ ii(m0 + 1) = ii(n0)

User applied rule cut formula

And closed the goal!

Rule argument

fma: n0 < m0

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m ≤ m; (s)% ⊦ ¬ n < n; (s)% ⊦ # [] = 0; (s)% ⊦ x to 0 = []; (s)% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n → m ≤ n; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s) Used formulas: n0 < m0