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Proof step 6 from proof from-from (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n + m ≤ 1 + # x0, ∀ n, m. n + m ≤ # x0 → x0 from m from n = x0 from m + n ⊦ a ' + x0 from m from n = a ' + x0 from m + n

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)