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Proof step 7 from proof infinite-append-butlast (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 8:
∀ n. σ[ii(n + 1) - 1] = lf(n + 1).first, ∀ n. n ≠ 0 → ii(n) ≠ 0, ii(0) = 0, ∀ n. ii(n + 1) = ii(n) + # (n = 0 ⊃ lf(0);lf(n).rest) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # (n = 0 ⊃ lf(0);lf(n).rest) = ⌜(n = 0 ⊃ lf(0);lf(n).rest)⌝ , ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n).last = lf(n + 1).first ∧ lf(n) ≠ [], (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).rest) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)) ⊦ (0 = 0 ⊃ 0;ii(0) - 1) = 0 ∧ (∀ n. (n + 1 = 0 ⊃ 0;ii(n + 1) - 1) = (n = 0 ⊃ 0;ii(n) - 1) + # lf(n) - 1 ∧ ⌜(n + 1 = 0 ⊃ 0;ii(n + 1) - 1)⌝ < # σ ∧ (σ from (n = 0 ⊃ 0;ii(n) - 1)) to # lf(n) = ⌜lf(n)⌝ ) ∧ (∀ m, n. m ≤ n → (m = 0 ⊃ 0;ii(m) - 1) ≤ (n = 0 ⊃ 0;ii(n) - 1)) ∧ (finp(σ) ⊃ ∃ n. # σ = ⌜(n = 0 ⊃ 0;ii(n) - 1)⌝ + ⌜1⌝ ∧ (∀ m. n ≤ m → lf(m).rest = [] ∧ (m = 0 ⊃ 0;ii(m) - 1) = (n = 0 ⊃ 0;ii(n) - 1)) ;∀ n. ∃ m. (m = 0 ⊃ 0;ii(m) - 1) ≤ n ∧ n < (m + 1 = 0 ⊃ 0;ii(m + 1) - 1) )

Rule argument

pos: right 1 terms: σ, λ n. (n = 0 ⊃ 0;ii(n) - 1) abort: F discard: T computed: F

Additional information

Used terms: σ, λ n. (n = 0 ⊃ 0;ii(n) - 1)