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Proof step 39 from proof infinite-append-butlast (previous step)

((n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1) + 1 = ii(n0 + 1), ii(0) = 0, ii(n0 + 1) = ii(n0) + # (n0 = 0 ⊃ lf(0);lf(n0).tail), ii(n1) = m0 + 1, (σ from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).tail) = ⌜(n0 = 0 ⊃ lf(0);lf(n0).tail)⌝ , m0 < ii(m) - 1, ⌜(n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1⌝ < # σ, ii(m) ≤ ii(n1), m ≤ n1, (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) ≠ [], ∀ 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 ⊦ ii(n1) = 0, m = 0, n1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 40:
((n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1) + 1 = ii(n0 + 1), ii(m) = 0, ii(n1) = m0 + 1, ii(n0 + 1) = ii(n0) + # (n0 = 0 ⊃ lf(0);lf(n0).tail), ii(0) = 0, (σ 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⌝ < # σ, m0 < 0 - 1, m ≤ n1, (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)⌝ , ∀ n. lf(n).last = lf(n + 1).head ∧ lf(n) ≠ [], ∀ n. σ[ii(n + 1) - 1] = lf(n + 1).head ⊦ m = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%leprf: ⊦ m ≤ m - 1 → m = 0; (forward) from specification nat% ⊦ m + 1 ≠ 0; (s)%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat