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Proof step 2 from proof firstn-app (previous step)

∀ n, y. firstn(# x0 + n, x0 + y) = x0 + firstn(n, y) ⊦ firstn(#(a + x0) + n, (a + x0) + y) = (a + x0) + firstn(n, y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ n, y. firstn(# x0 + n, x0 + y) = x0 + firstn(n, y) ⊦ firstn(# x0 + n, x0 + y) = x0 + firstn(n, y)

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.head + firstn(n, x.tail);% ⊦ x + y = x + z ↔ y = z; (s)%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ (a ' + x).tail = x; (s)% ⊦ (a ' + x).head = a; (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)