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Proof step 1 from proof infinite-append-butlast (conclusion)

∀ n. lf(n).last = lf(n + 1).first ∧ lf(n) ≠ [] ⊦ ∃ 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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ n. lf(n).last = lf(n + 1).first ∧ lf(n) ≠ [] ⊦ ∃ 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))

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf