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Proof step 40 from proof infinite-append-lem (previous step)

ii(m) + n = n1 + ii(n0), ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), m < n0, n < # lf(m), n1 < # lf(n0), ∀ 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) ⊦ lf(n0)[n1] = lf(m)[n], lf(m) = [], lf(n0) = [], m = n0 + 1, n0 = m

User applied rule all left

And obtained the following new goals

Proof step 41:
ii(m) + n = n1 + ii(n0), ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, (nf from ii(m)) to # lf(m) = lf(m), n1 < # lf(n0), n < # lf(m), m < n0, n0 - 1 < n0 → ii((n0 - 1) + 1) = ii(n0 - 1) + # lf(n0 - 1) ∧ (nf from ii(n0 - 1)) to # lf(n0 - 1) = lf(n0 - 1) ∧ (∀ m0. m0 ≤ n0 - 1 → ii(m0) ≤ ii(n0 - 1)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(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)) ⊦ lf(n0)[n1] = lf(m)[n], lf(n0) = [], lf(m) = [], m = n0 + 1, n0 = m

Rule argument

pos: left 8 terms: n0 - 1 abort: F discard: F computed: F

Additional information

Used terms: n0 - 1