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Proof step 132 from proof infinite-append-butlast (previous step)

ii(n1 + 1) = m0 + 1, ii(0) = 0, ii(2 + n1) = 2 + m0 + # y, lf(2 + n1) = a ' + x1, (σ from m0 + 1) to # y + 1 = ⌜y⌝ + ⌜a '⌝ , (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, ∀ m, n. m ≤ n → ii(m) ≤ ii(n) ⊦ σ[m0 + # y + 1] = a, # σ < ⌜2⌝ + ⌜m0⌝ + ⌜# y⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: stream name: to-unwind-end seq: ⌜n⌝ < # σ ⊦ σ to n + 1 = (σ to n) + ⌜σ[n] '⌝ subst: [n, σ] → [# y, σ from m0 + 1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%lele-trans: ⊦ ¬ n∞1 < n∞ ∧ ¬ n∞2 < n∞1 → ¬ n∞2 < n∞; (forward) from specification nat-inf% ⊦ ⌜n⌝ + n∞1 < ⌜n⌝ + n∞2 ↔ n∞1 < n∞2; (s)% ⊦ ⌜1⌝ < n∞ ↔ n∞ ≠ ⌜0⌝ ∧ n∞ ≠ ⌜1⌝ ; (s)% ⊦ n < m → (⌜(* n)⌝ = ⌜(* m)⌝ + n∞2 ↔ false); (s)%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%⌜n⌝ < # σ ⊦ ⌜n0⌝ < #(σ from n) ↔ ⌜n0 + n⌝ < # σ; (s)% ⊦ ⌜m⌝ = ⌜m0⌝ ↔ m = m0; (s) Used terms: # y, σ from m0 + 1