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Proof step 61 from proof infinite-append-lem (previous step)

ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), m0 ≤ m, m < n0, ii(m) < (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)), m0 < 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) ⊦ m = n0 + 1, n0 = m, n0 = 0, (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)) = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 62:
m0 ≠ n0 + 1, ii(m) < ii(m0), ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), m0 ≤ m, m < n0, m0 < 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) ⊦ m = n0 + 1, n0 = m, n0 = 0, (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)) = 0

Rule argument

pos: left 6

Additional information

Used simplifier rules: succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2