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Proof step 17 from proof from-append-add (previous step)

σ = ⌜nf⌝ ⊦ # σ < ⌜n⌝ , (σ to n) + ⌜x⌝ from # x + n = [] ∧ (⌜n0⌝ < #((σ to n) + ⌜x⌝ from # x + n) → ((σ to n) + ⌜x⌝ from # x + n)[n0] = ([])[n0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
⊦ (nf to n) + x from # x + n = []

Rule argument

Additional information

Used simplifier rules: size-inf: ⊦ # ⌜nf⌝ = ∞;%to-inf: ⊦ ⌜nf⌝ to n = ⌜nf to n⌝ ;%from-appfin: ⊦ ⌜x⌝ + ⌜y⌝ from n = ⌜x + y from n⌝ ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%fin-empty: ⊦ ⌜x⌝ = [] ↔ x = [];%size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ;%size-inf-to: ⊦ #(nf to n) = n;%length-from: ¬ # x < n ⊦ #(x from n) = # x - n;%zero-01: ⊦ ¬ m < 0; from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%zero: ⊦ m - m = 0; from specification nat%cancel-01: ⊦ m + n - n + n1 = m - n1; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%less-inf1: ⊦ ¬ ∞ < n∞; from specification nat-inf