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Proof step 8 from proof get-append-less (previous step)

n = m + 1, n ≤ # x0, ∀ n, y. n < # x0 → (x0 + y)[n] = x0[n] ∨ x0 = [] ⊦ n = 0, (x0 + y)[m] = (a ' + x0)[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
m < # x0, ∀ n, y0. n < # x0 → (x0 + y0)[n] = x0[n] ⊦ (x0 + y)[m] = x0[m], x0 = []

Rule argument

Additional information

Used simplifier rules: get-succ: ⊦ (a ' + x)[n + 1] = x[n];%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2