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Proof step 12 from proof put-at (previous step)

m = m0 + 1, m ≤ # x0, ∀ m, n, a. m < # x0 → m ≠ n → x0[m, a][n] = x0[n] ⊦ m = 0, m = n, (a0 ' + x0[m0, a])[n] = (a0 ' + x0)[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
m0 < # x0, ∀ m, n, a. m < # x0 → m ≠ n → x0[m, a][n] = x0[n] ⊦ (a0 ' + x0[m0, a])[n] = (a0 ' + x0)[n], m0 + 1 = n, x0 = []

Rule argument

Additional information

Used simplifier rules: 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%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