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Proof step 11 from proof put-at (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
m ≤ # x0, ∀ m, a, n. m < # x0 → x0[m, a][n] = x0[n] ∨ m = n ⊦ (a0 ' + x0[m - 1, a])[n] = (a0 ' + x0)[n], m = 0, m = n

Rule argument

Additional information

Used simplifier rules: ⊦ a ' + x ≠ []; (s)% ⊦ n -1 = n - 1; (s)% ⊦ (a ' + x).tail = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).head = a; (s)