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Proof step 16 from proof infinite-append-lem (previous step)

ii(0) = 0, ii(m) < (m0 = m + 1 ⊃ ii(m) + # lf(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)) ⊦

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 17:
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

Rule argument

pos: left 2

Additional information

Used simplifier rules: ⊦ m + m0 ≤ m ↔ m0 = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic