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Proof step 51 from proof infinite-append-lem (previous step)

ii(m) + n < ii(m1), (nf from ii(m1)) to # lf(m1) = lf(m1), ii(m1 + 1) = ii(m1) + # lf(m1), (nf from ii(m)) to # lf(m) = lf(m), ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, ii(m) + n = n1 + ii(m1 + 1), n1 < # lf(m1 + 1), n < # lf(m), m < m1, ii(m + 1) ≤ ii(m1), ∀ m0. ¬ m1 < m0 → ii(m0) ≤ ii(m1), ∀ m2. ¬ m1 < m2 → ii(m2 + 1) = ii(m2) + # lf(m2) ∧ (nf from ii(m2)) to # lf(m2) = lf(m2) ∧ (∀ m0. m0 ≤ m2 → ii(m0) ≤ ii(m2)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m) ⊦ m1 = m, m = 2 + m1, lf(m) = [], ii(m1 + 1) + # lf(m1 + 1) = 0, lf(m1 + 1) = [], lf(m1 + 1)[n1] = lf(m)[n], m1 + 1 = m, m1 = 0

User applied rule insert equation

And obtained the following new goals

Rule argument

pos: left 7 flags: discard

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