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Proof step 131 from proof infinite-append-butlast (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: less-size-from: ⌜n⌝ < # σ ⊦ ⌜n0⌝ < #(σ from n) ↔ ⌜n0 + n⌝ < # σ; from specification stream%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%ls-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; from specification nat-inf%less-succ-leq: ⊦ ⌜n⌝ < ⌜1⌝ + n∞ ↔ ¬ n∞ < ⌜n⌝ ; from specification nat-inf%add-ls-cancel-num-04: ⊦ m ≤ n → (⌜(* m)⌝ + n∞2 < ⌜(* n)⌝ + n∞1 ↔ n∞2 < ⌜n - m⌝ + n∞1); from specification nat-inf%lele-trans: ⊦ ¬ n∞1 < n∞ ∧ ¬ n∞2 < n∞1 → ¬ n∞2 < n∞; (forward) from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf