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Proof step 5 from proof from-from (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n + m ≤ # x0 + 1, ∀ 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%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