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Proof step 17 from proof put-at (previous step)

n = m1 + 1, m0 < # x0, ∀ m, n, a. m < # x0 → m ≠ n → x0[m, a][n] = x0[n] ⊦ m0 + 1 = n, x0 = [], n = 0, x0[m0, a][m1] = (a0 ' + x0)[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
m0 < # x0, ∀ m, n, a. m < # x0 → m ≠ n → x0[m, a][n] = x0[n] ⊦ m0 = m1, x0 = [], x0[m0, a][m1] = x0[m1]

Rule argument

Additional information

Used simplifier rules: get-succ: ⊦ (a ' + x)[n + 1] = x[n];%a: ⊦ (x + y) + z = x + y + z; from specification list%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2