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Proof step 4 from proof blastN-firstN (previous step)

n ≤ #(a + x0), ∀ n. n ≤ # x0 → butlastn(n, x0) = firstn(# x0 - n, x0) ⊦ butlastn(n, a + x0) = firstn(#(a + x0) - n, a + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n ≤ 1 + # x0, ∀ n. n ≤ # x0 → butlastn(n, x0) = firstn(# x0 - n, x0) ⊦ butlastn(n, a ' + x0) = firstn(1 + # x0 - n, 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)