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Proof step 3 from proof from-succ (previous step)

⌜n⌝ < # ⌜x⌝ ⊦ ⌜x⌝ from n = ⌜⌜x⌝ [n] '⌝ + (⌜x⌝ from n + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n < # x ⊦ x[n] ' + (x from n + 1) = x from n, x = []

Rule argument

Additional information

Used simplifier rules: from-fin: ⊦ ⌜x⌝ from n = ⌜x from n⌝ ;%sel-fin: ⊦ ⌜x⌝ [n] = x[n];%eq-stream-to-list-2-1: ⊦ ⌜x⌝ + ⌜y⌝ = ⌜z⌝ ↔ x + y = z;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ;%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf