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Proof step 1 from proof union-subset-05 (conclusion)

⊦ s ⊆ s0 → s ∪ s1 ⊆ (s0 ∪ s1) ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a0 ∈ s, a1 ∈ s, ∀ a. a ∈ s → a ∈ s0 ⊦ a0 = a, a0 ∈ s1, a0 ∈ s0

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-empty: ⊦ ¬ a ∈ ∅; from specification set-basic%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%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