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

#(x ∪ y ∪ z) ≤ #(x ∪ z) ⊦ y ⊆ x ∪ z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# rmdup(x + y + z) = # rmdup(x + z) ⊦ y ⊆ x + z

Rule argument

Additional information

Used simplifier rules: union: ⊦ x ∪ y = rmdup(x + y);%assoc: ⊦ (x ∪ y) ∪ z = x ∪ y ∪ z;%rmdup-01: ⊦ x ⊆ rmdup(y) ↔ x ⊆ y;% ⊦ m ≤ m; (s)%rm-size-04: ⊦ # rmdup(x + y + z) ≤ # rmdup(x + z) → # rmdup(x + y + z) = # rmdup(x + z); (forward) from specification list-dup%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ rmdup(rmdup(x) + y) = rmdup(x + y); (s)