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Proof step 20 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, m = 0

User applied rule all left

And obtained the following new goals

Proof step 21:
ii(0) = 0, ii(m) < ii(m0), m0 < m, m - 1 < m → ii((m - 1) + 1) = ii(m - 1) + # lf(m - 1) ∧ (nf from ii(m - 1)) to # lf(m - 1) = lf(m - 1) ∧ (∀ m1. m1 ≤ m - 1 → ii(m1) ≤ ii(m - 1)), ∀ 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

pos: left 4 terms: m - 1 abort: F discard: F computed: F

Additional information

Used terms: m - 1