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Proof step 7 from proof put-app-left (previous step)

n = m + 1, n ≤ # x0, ∀ n, y, a. n < # x0 → (x0 + y)[n, a] = x0[n, a] + y ⊦ n = 0, a0 ' + (x0 + y)[m, a] = (a0 ' + x0)[n, a] + y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
m < # x0, ∀ n, y, a. n < # x0 → (x0 + y)[n, a] = x0[n, a] + y ⊦ (x0 + y)[m, a] = x0[m, a] + y, x0 = []

Rule argument

Additional information

Used simplifier rules: put-succ: ⊦ (b ' + x)[n + 1, a] = b + x[n, a];%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%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; 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