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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.first + firstn(n, x.rest);%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%c-08: ⊦ (a ' + x).first = a; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list