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Proof step 3 from proof restn-empty (previous step)

n ≤ #(a + x0), ∀ n. n ≤ # x0 → (restn(n, x0) = [] ↔ n = # x0) ⊦ restn(n, a + x0) = [] ↔ n = #(a + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
restn(n, a ' + x0) = [], n ≤ # x0, ∀ n0. n0 < # x0 → restn(n0, x0) = [] → n0 = # x0 ⊦ n = 1 + # x0

Rule argument

Additional information

Used simplifier rules: restN-rec: ⊦ restn(n + 1, x) = restn(n, x.tail);%restN-all: ⊦ restn(# x, x) = [];% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ (a ' + x).tail = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)