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Proof step 59 from proof infinite-append-prefixnextli (previous step)

ii(0) = 0, lf(0) = [], ii(n0 + 1) = ii(n0) + # nextli0(lf(n0)), (σ from ii(n0)) to # nextli0(lf(n0)) = ⌜nextli0(lf(n0))⌝ , ii(n0 + 1) ≤ ii(n0), ∀ n0. lf(n0) = (σ to ii(n0)).l, ∀ n. ii(n + 1) = ii(n) + # nextli0(lf(n)) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # nextli0(lf(n)) = ⌜nextli0(lf(n))⌝ , ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextli(lf(n)), ∀ x. nextli(x) = x + nextli0(x), (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → nextli0(lf(m)) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)) ⊦ # σ = ⌜ii(n0)⌝ , nextli0(lf(n0)) = [], # σ < ⌜ii(n0 + 1)⌝

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 3 flags: discard

Additional information

Used simplifier rules: le-add: ⊦ m + m0 ≤ m ↔ m0 = 0; from specification nat%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