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Proof step 39 from proof infinite-append-prefixnextli (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 40:
lf(0) = [], ii(0) = 0, nextli0(lf(n0)) = [], ii(n0 + 1) = ii(n0), ii(m) ≤ ii(n0) + 1, ii(n0) + 1 < ii(m + 1), ii(m) < ii(m + 1), ∀ m. n0 < m → ii(m) = ii(n0) ∧ lf(m) = lf(n0), ∀ x. nextli(x) = x + nextli0(x), ∀ n. lf(n + 1) = nextli(lf(n)), m + 1 ≤ n0 → ii(m + 1) ≤ ii(n0), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n) = nf to ii(n), ∀ n. ii(n + 1) = ii(n) + # nextli0(lf(n)) ∧ (nf from ii(n)) to # nextli0(lf(n)) = nextli0(lf(n)) ⊦ ii(m + 1) = 0

Rule argument

pos: left 11 terms: m + 1, n0 abort: F discard: F computed: F

Additional information

Used terms: m + 1, n0