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Proof step 1 from proof sub-05 (conclusion)

⊦ x ∪ x0 ⊆ y ∪ z ∪ x ↔ x0 ⊆ y ∪ z ∪ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ x0, ∀ a. a ∈ x0 → a ∈ y ∨ a ∈ z ∨ a ∈ x ⊦ a ∈ y, a ∈ z, a ∈ x

Rule argument

Additional information

Used simplifier rules: union: ⊦ x ∪ y = rmdup(x + y);%assoc: ⊦ (x ∪ y) ∪ z = x ∪ y ∪ z;%rmdup: ⊦ rmdup(x) ⊆ y ↔ x ⊆ y;%rmdup-01: ⊦ x ⊆ rmdup(y) ↔ x ⊆ y;%subset: ⊦ x ⊆ y ↔ (∀ a. a ∈ x → a ∈ y);% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ rmdup(rmdup(x) + y) = rmdup(x + y); (s)