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Proof step 8 from proof from-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n ≤ # x0, ∀ n. n < # x0 → x0[n] ' + (x0 from n + 1) = x0 from n ∨ x0 = [] ⊦ n = 0, x0[n - 1] ' + (x0 from n) = a ' + x0 from n

Rule argument

Additional information

Used simplifier rules: pred-03: ⊦ n -1 = n - 1; from specification nat