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Proof step 48 from proof infinite-append-butlast (previous step)

# x = ii(n), ((n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1) + 1 = ii(n0 + 1), ii(n0 + 1) = ii(n0) + # (n0 = 0 ⊃ lf(0);lf(n0).tail), ii(0) = 0, (x from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).tail) = (n0 = 0 ⊃ lf(0);lf(n0).tail), (n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1 < ii(n), ii(n0 + 1) ≤ ii(n), n ≤ m, ∀ n1. ii(n1 + 1) = ii(n1) + # (n1 = 0 ⊃ lf(0);lf(n1).tail) ∧ ¬ ii(n) < ii(n1 + 1) ∧ (x from ii(n1)) to # (n1 = 0 ⊃ lf(0);lf(n1).tail) = (n1 = 0 ⊃ lf(0);lf(n1).tail), ∀ n1. x[ii(n1 + 1) - 1] = lf(n1 + 1).head, ∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).tail) = [] ∧ ii(m) = ii(n), ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n).last = lf(n + 1).head ∧ lf(n) ≠ [] ⊦ lf(m).tail = [] ∧ (m = 0 ⊃ 0;ii(m) - 1) = (n = 0 ⊃ 0;ii(n) - 1)

User applied rule all left

And obtained the following new goals

Proof step 49:
# x = ii(n), ((n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1) + 1 = ii(n0 + 1), lf(m).last = lf(m + 1).head, ii(0) = 0, ii(n0 + 1) = ii(n0) + # (n0 = 0 ⊃ lf(0);lf(n0).tail), (x from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).tail) = (n0 = 0 ⊃ lf(0);lf(n0).tail), (n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1 < ii(n), n ≤ m, ii(n0 + 1) ≤ ii(n), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).tail) = [] ∧ ii(m) = ii(n), ∀ n1. x[ii(n1 + 1) - 1] = lf(n1 + 1).head, ∀ n1. ii(n1 + 1) = ii(n1) + # (n1 = 0 ⊃ lf(0);lf(n1).tail) ∧ ¬ ii(n) < ii(n1 + 1) ∧ (x from ii(n1)) to # (n1 = 0 ⊃ lf(0);lf(n1).tail) = (n1 = 0 ⊃ lf(0);lf(n1).tail) ⊦ lf(m) = [], lf(m).tail = [] ∧ (m = 0 ⊃ 0;ii(m) - 1) = (n = 0 ⊃ 0;ii(n) - 1)

Rule argument

pos: left 14 terms: m abort: F discard: T computed: F

Additional information

Used terms: m