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Proof step 1 from proof size-nle (conclusion)

⊦ ¬ y ⊆ x ∪ z → ¬ #(x ∪ y ∪ z) ≤ #(x ∪ z)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# 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;%ler: ⊦ m ≤ m; from specification nat%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%rm-rm-01: ⊦ rmdup(rmdup(x) + y) = rmdup(x + y); from specification list-dup