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Proof step 11 from proof from-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
m < # x0, ∀ n. n < # x0 → x0[n] ' + (x0 from n + 1) = x0 from n ⊦ x0[m] ' + (x0 from m + 1) = x0 from m, x0 = []

Rule argument

Additional information

Used simplifier rules: from-succ-cons-l-lem: ⊦ a ' + x from n + 1 = x from n;% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)