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Proof step 6 from proof sublist-append-01 (previous step)

∀ m, n0, y. firstn(m, restn(n0 + # x0, x0 + y)) = firstn(m, restn(n0, y)) ⊦ firstn(m, restn(n0 + #(a + x0), (a + x0) + y)) = firstn(m, restn(n0, y))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ m, n0, y. firstn(m, restn(n0 + # x0, x0 + y)) = firstn(m, restn(n0, y)) ⊦ firstn(m, restn(n0 + # x0, x0 + y)) = firstn(m, restn(n0, y))

Rule argument

Additional information

Used simplifier rules: restN-rec: ⊦ restn(n + 1, x) = restn(n, x.tail);%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ (a ' + x).tail = x; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)