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Proof step 53 from proof infinite-append-prefixnextli (previous step)

lf(m0) = lf(n0), ii(m0) = ii(n0), lf(0) = [], ii(0) = 0, nextli0(lf(n0)) = [], ii(n0 + 1) = ii(n0), ii(m0 + 1) = ii(n0), lf(m0 + 1) = nextli(lf(n0)), n0 < m0, (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → nextli0(lf(m)) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)), ∀ x. nextli(x) = x + nextli0(x), ∀ n. lf(n + 1) = nextli(lf(n)), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. ii(n + 1) = ii(n) + # nextli0(lf(n)) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # nextli0(lf(n)) = ⌜nextli0(lf(n))⌝ , ∀ n0. lf(n0) = (σ to ii(n0)).l ⊦ # σ = ⌜ii(n0)⌝ , m0 = 0, nextli(lf(n0)) = lf(n0), # σ < ⌜ii(n0)⌝

User applied rule all left

And closed the goal!

Rule argument

pos: left 11 terms: lf(n0) abort: F discard: F computed: T

Additional information

Used simplifier rules: app: ⊦ x + [] = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: lf(n0)