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Proof step 6 from proof put-length (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n ≤ # x0, ∀ n, a. n < # x0 → # x0[n, a] = # x0 ∨ x0 = [] ⊦ n = 0, # x0[n - 1, a] = # x0

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ []; from specification list%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%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%pred-03: ⊦ n -1 = n - 1; from specification nat%c-07: ⊦ (a ' + x).rest = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-08: ⊦ (a ' + x).first = a; from specification list