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Proof step 5 from proof restn-less (previous step)

∀ y, a. n0 ≤ # y → restn(n0, a ' + y) = (a ' + y)[n0] ' + restn(n0, y), n0 +1 ≤ # y ⊦ restn(n0 +1, a ' + y) = (a ' + y)[n0 +1] ' + restn(n0 +1, y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: restN-rec: ⊦ restn(n + 1, x) = restn(n, x.rest);%get-succ: ⊦ (a ' + x)[n + 1] = x[n];%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2