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Proof step 60 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), m < 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) ⊦ m0 ≤ m → (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)) ≤ ii(m), m = n0 + 1, n0 = m, n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 61:
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

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat