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Proof step 1 from proof suband (conclusion)

⊦ x ∪ y ⊆ z ↔ x ⊆ z ∧ y ⊆ z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (∀ a. a ∈ x ∨ a ∈ y → a ∈ z) ↔ (∀ a. a ∈ x → a ∈ z) ∧ (∀ a. a ∈ y → a ∈ z)

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;%subset: ⊦ x ⊆ y ↔ (∀ a. a ∈ x → a ∈ y);%a: ⊦ (x + y) + z = x + y + z; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list