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

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

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
m + n0 +1 < # x, ∀ m, x. m + n0 < # x → restn(n0, x)[m] = x[m + n0] ∨ x = [] ⊦ restn(n0 +1, x)[m] = x[m + n0 +1], x = []

Rule argument

term: n

Additional information

Used simplifier rules: restN-zero: ⊦ restn(0, x) = x;%nat freely generated by 0, +1;;% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic