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Proof step 16 from proof put-at (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
m0 < # x0, ∀ m, a, n. m < # x0 → x0[m, a][n] = x0[n] ∨ m = n ⊦ m0 + 1 = n, x0[m0, a][n - 1] = (a0 ' + x0)[n], n = 0, x0 = []

Rule argument

Additional information

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