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Proof step 89 from proof infinite-append-butlast (previous step)

ii(0) = 0, (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).tail) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)), ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ n. ii(n + 1) = ii(n) + # (n = 0 ⊃ lf(0);lf(n).tail) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # (n = 0 ⊃ lf(0);lf(n).tail) = ⌜(n = 0 ⊃ lf(0);lf(n).tail)⌝ , ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n).last = lf(n + 1).head ∧ lf(n) ≠ [] ⊦ σ[ii(1) - 1] = lf(1).head

User applied rule all left

And obtained the following new goals

Proof step 90:
ii(0) = 0, ii(0 + 1) = ii(0) + # (0 = 0 ⊃ lf(0);lf(0).tail), (σ from ii(0)) to # (0 = 0 ⊃ lf(0);lf(0).tail) = ⌜(0 = 0 ⊃ lf(0);lf(0).tail)⌝ , (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).tail) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)), ∀ n. lf(n).last = lf(n + 1).head ∧ lf(n) ≠ [], ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. ii(n + 1) = ii(n) + # (n = 0 ⊃ lf(0);lf(n).tail) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # (n = 0 ⊃ lf(0);lf(n).tail) = ⌜(n = 0 ⊃ lf(0);lf(n).tail)⌝ , ∀ n. n ≠ 0 → ii(n) ≠ 0 ⊦ σ[ii(1) - 1] = lf(1).head, # σ < ⌜ii(0 + 1)⌝

Rule argument

pos: left 4 terms: 0 abort: F discard: F computed: F

Additional information

Used terms: 0