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Proof step 15 from proof put-at (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
m0 < # x0, ∀ m, n, a. m < # x0 → m ≠ n → x0[m, a][n] = x0[n] ⊦ m0 + 1 = n, x0 = [], n = 0, x0[m0, a][n - 1] = (a0 ' + x0)[n]

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list