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Proof step 23 from proof infinite-append-butlast (previous step)

(m0 + # lf(n0) - 1) + 1 = ii(n0 + 1), ii(n0 + 1) = ii(n0) + # lf(n0).tail, ii(0) = 0, ii(n0) = m0 + 1, (σ from ii(n0)) to # lf(n0).tail = ⌜lf(n0).tail⌝ , ⌜m0 + # lf(n0) - 1⌝ < # σ, (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) ≠ [], ∀ n. σ[ii(n + 1) - 1] = lf(n + 1).head ⊦ ii(n0) = 0, n0 = 0, (σ from m0) to # lf(n0) = ⌜lf(n0)⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
(# lf(n0).tail + m0) + 1 = ii(n0 + 1), m0 + # lf(n0) - 1 = # lf(n0).tail + m0, ii(n0) = m0 + 1, ii(0) = 0, (σ from m0 + 1) to # lf(n0).tail = ⌜lf(n0).tail⌝ , ⌜# lf(n0).tail⌝ + ⌜m0⌝ < # σ, (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. σ[ii(n + 1) - 1] = lf(n + 1).head, ∀ 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 from m0) to # lf(n0) = ⌜lf(n0)⌝

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)% ⊦ m + 1 ≠ 0; (s)%eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic