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

ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), m < n0, n < # lf(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ n1. n1 < # lf(m) → nf(n1 + ii(m)) = lf(m)[n1], ¬ ii(m) + n < ii(n0) + # lf(n0) ∨ ii(m) + n < ii(n0) ⊦ lf(m) = [], m = n0 + 1, n0 = m, nf(ii(m) + n) = lf(m)[n]

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: n

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic Used terms: n