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Proof step 4 from proof sublist-part (previous step)

#(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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
1 + # 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

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)