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Proof step 61 from proof infinite-append-lem (previous step)

ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, (nf from ii(m)) to # lf(m) = lf(m), m0 < n0, ii(m) < (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)), m < n0, m0 ≤ m, ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦ m = n0 + 1, n0 = m

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 62:
ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), m < n0, m0 < n0, ii(m) < ii(m0), m0 ≤ m, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m) ⊦ m = n0 + 1, m0 = n0 + 1, n0 = m

Rule argument

pos: left 5

Additional information

Used simplifier rules: ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic