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

ii(n0) = m0 + 1 ∧ ii(n0) ≠ 0, ii(0) = 0, (σ from ii(n0)) to # lf(n0).rest = ⌜lf(n0).rest⌝ , ii(n0 + 1) = ii(n0) + # lf(n0).rest, ii(n0 + 1) = (m0 + # lf(n0) - 1) + 1, ⌜m0 + # lf(n0) - 1⌝ < # σ, ∀ 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, 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 from m0) to # lf(n0) = ⌜lf(n0)⌝ , n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2