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Proof step 41 from proof infinite-append-butlast (previous step)

ii(n1) = m1 + 1 ∧ ii(n1) ≠ 0, (σ from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).rest) = ⌜(n0 = 0 ⊃ lf(0);lf(n0).rest)⌝ , m0 + 1 = ii(n0) + # (n0 = 0 ⊃ lf(0);lf(n0).rest), ii(n0 + 1) = m0 + 1, ii(0) = 0, (n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) = m0 + 1, m1 < ii(m) - 1, ii(m) ≤ ii(n1), m ≤ n1, ⌜m0⌝ < # σ, ∀ 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)), ∀ n. σ[ii(n + 1) - 1] = lf(n + 1).first, ∀ n. n ≠ 0 → ii(n) ≠ 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 = 0, 0 = ii(m) - 1, n1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: lez: ⊦ 0 ≤ m; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%leprf: ⊦ m ≤ m - 1 → m = 0; (forward) from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat