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Proof step 18 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 + 1, ii(m0) = 0

User applied rule cut formula

And obtained the following new goals

Proof step 19:
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

Rule argument

fma: m0 = m

Additional information

Used simplifier rules: eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used formulas: m0 = m