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Proof step 1 from proof union-subset-04 (conclusion)

⊦ s ⊆ s0 ∧ s1 ⊆ s2 ∧ s3 ⊆ s4 → s ∪ s1 ∪ s3 ⊆ (s0 ∪ s2 ∪ s4) ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2, ∀ a. a ∈ s3 → a ∈ s4, a0 ∈ s ∨ a0 ∈ s1 ∨ a0 ∈ s3 ⊦ a0 = a, a0 ∈ s0, a0 ∈ s2, a0 ∈ s4

Rule argument

Additional information

Used simplifier rules: unrax: ⊦ a ∈ s ∪ s0 ↔ ¬ ¬(a ∈ s ∨ a ∈ s0);%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; from specification set-basic%Subset: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); from specification set-basic