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Proof step 2 from proof len (previous step)

∀ y. #(x0 + y) = # x0 + # y ⊦ #((a + x0) + y) = #(a + x0) + # y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ y. #(x0 + y) = # x0 + # y ⊦ #(x0 + y) = # x0 + # y

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%len-05: ⊦ #(a ' + x) = (# x)+1;%c: ⊦ m + n = n + m; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2