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Proof step 45 from proof infinite-append-lem (previous step)

(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), m ≤ m1, n1 < # lf(m1 + 1), n < # lf(m), ¬ m1 < m + 1 → 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) ⊦ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
(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), m ≤ m1, n1 < # lf(m1 + 1), n < # lf(m), ∀ 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), m < m1 → ii(m + 1) ≤ ii(m1) ⊦ 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

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2