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

m + n ≤ # x ⊦ lastn(m, restn(n, x)) = lastn(m, x)

User applied rule structural induction

And obtained the following new goals

Proof step 4:
m + n ≤ #(a + x0), ∀ m, n. m + n ≤ # x0 → lastn(m, restn(n, x0)) = lastn(m, x0) ⊦ lastn(m, restn(n, a + x0)) = lastn(m, a + x0)

Rule argument

indvar: x

Additional information

Used simplifier rules: restN-zero: ⊦ restn(0, x) = x;%lastN-zero: ⊦ lastn(0, x) = [];%list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic