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

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 17:
m0 ≠ m + 1, ii(m) < ii(m0), ii(0) = 0, 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(m) + # lf(m);ii(m0)) = 0

Rule argument

pos: left 3

Additional information

Used simplifier rules: l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%ref-1: ⊦ ¬ n < n; from specification nat-basic1