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Proof step 12 from proof infinite-append-lem (previous step)

ii(0) = 0, ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)) ⊦ ((λ n. (n < ii(m) + # lf(m) ∧ ¬ n < ii(m) ⊃ lf(m)[n - ii(m)];nf(n))) from ii(m)) to # lf(m) = lf(m) ∧ (m0 ≤ m → (m0 = m + 1 ⊃ ii(m) + # lf(m);ii(m0)) ≤ ii(m))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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