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Proof step 4 from proof append-from-le (previous step)

n ≤ #(a + x0), ∀ n, y. n ≤ # x0 → x0 + y from n = (x0 from n) + y ⊦ (a + x0) + y from n = (a + x0 from n) + y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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