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Proof step 7 from proof put-length (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n ≤ # x0, ∀ n, a. n < # x0 → # x0[n, a] = # x0 ∨ x0 = [] ⊦ # x0[n - 1, a] = # x0, n = 0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ n -1 = n - 1; (s)% ⊦ (a ' + x).tail = x; (s)% ⊦ (a ' + x).head = a; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)