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Proof step 19 from proof infinite-append-lem (previous step)

ii(0) = 0, 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)) ⊦ m0 = m, m0 = m + 1, ii(m0) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
ii(0) = 0, 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)) ⊦ m0 = m, m0 = m + 1, ii(m0) = 0, m = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat