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Proof step 21 from proof infinite-append-lem (previous step)

ii(0) = 0, ii(m) < ii(m0), m0 < m, m - 1 < m → ii((m - 1) + 1) = ii(m - 1) + # lf(m - 1) ∧ (nf from ii(m - 1)) to # lf(m - 1) = lf(m - 1) ∧ (∀ m1. m1 ≤ m - 1 → ii(m1) ≤ ii(m - 1)), ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)) ⊦ m0 = m, m0 = m + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
ii(0) = 0, ii((m - 1) + 1) = ii(m - 1) + # lf(m - 1), (nf from ii(m - 1)) to # lf(m - 1) = lf(m - 1), m0 < m, ii(m) < ii(m0), ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)), ∀ m1. m1 < m → ii(m1) ≤ ii(m - 1) ⊦ m0 = m + 1, m0 = m

Rule argument

Additional information

Used simplifier rules: ⊦ m + 1 ≤ n ↔ m < n; (s)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)