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Proof step 1 from proof sub-04 (conclusion)

⊦ x ∪ x0 ⊆ y ∪ x ∪ z ↔ x0 ⊆ y ∪ x ∪ z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ x0, ∀ a. a ∈ x0 → a ∈ y ∨ a ∈ x ∨ a ∈ z ⊦ a ∈ y, a ∈ x, 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;%rmdup-01: ⊦ x ⊆ rmdup(y) ↔ x ⊆ y;%subset: ⊦ x ⊆ y ↔ (∀ a. a ∈ x → a ∈ y);%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%rm-rm-01: ⊦ rmdup(rmdup(x) + y) = rmdup(x + y); from specification list-dup