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Proof step 2 from proof size-nls (previous step)

#((x ∪ y) ∪ z) < #(x ∪ y) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# rmdup((x + y) + z) < # rmdup(x + y) ⊦ x + y = []

Rule argument

Additional information

Used simplifier rules: union: ⊦ x ∪ y = rmdup(x + y);%assoc: ⊦ (x ∪ y) ∪ z = x ∪ y ∪ z;% ⊦ rmdup(x) = [] ↔ x = []; (s)% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ rmdup(rmdup(x) + y) = rmdup(x + y); (s)