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Proof step 2 from proof restn-less (previous step)

n < # x ⊦ restn(n, x) = x[n] + restn(n +1, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < # x ⊦ restn(n, x) = x[n] ' + restn(n, x.tail), x = []

Rule argument

Additional information

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