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Proof step 23 from proof infinite-append-lem (previous step)

m = m2 + 1, ii(0) = 0, ii(m2 + 1) = ii(m2) + # lf(m2), (nf from ii(m2)) to # lf(m2) = lf(m2), ii(m) < ii(m0), m0 < m, ∀ 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(m2) ⊦ m0 = m, m0 = m + 1, ii(m0) = 0, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
(nf from ii(m2)) to # lf(m2) = lf(m2), ii(m2 + 1) = ii(m2) + # lf(m2), ii(0) = 0, m0 ≤ m2, ii(m2 + 1) < ii(m0), ∀ m. ¬ m2 < m → ii(m) ≤ ii(m2), ∀ m. ¬ m2 < m → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m1. m1 ≤ m → ii(m1) ≤ ii(m)) ⊦ m0 = 2 + m2, m0 = m2 + 1, ii(m0) = 0

Rule argument

Additional information

Used simplifier rules: lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2