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Proof step 3 from proof sublist-part (previous step)

# x = n0 + m0 ⊦ firstn(m0, x) + firstn(n0, restn(m0, x)) = x

User applied rule structural induction

And obtained the following new goals

Proof step 4:
#(a + x0) = n0 + m0, ∀ n0, m0. # x0 = n0 + m0 → firstn(m0, x0) + firstn(n0, restn(m0, x0)) = x0 ⊦ firstn(m0, a + x0) + firstn(n0, restn(m0, a + x0)) = a + x0

Rule argument

indvar: x

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];%restN-zero: ⊦ restn(0, x) = x;%list freely generated by [], +;;% ⊦ # [] = 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% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list