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Proof step 2 from proof insert-subset-old (previous step)

ks ⊆ ks0 ⊦ ks ++ key ⊆ ks0 ∪ ks1 ↔ key ∈ ks0 ∨ key ∈ ks1

User applied rule case distinction

And obtained the following new goals

Proof step 3:
ks ⊆ ks0, key ∈ ks0 ∨ key ∈ ks1 ⊦ ks ++ key ⊆ ks0 ∪ ks1

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ ks ++ key ⊆ ks0 ↔ ks ⊆ ks0 ∧ key ∈ ks0; (ws)% ⊦ ks ⊆ ks0 → ks ⊆ ks0 ∪ ks1; (s)%una: ⊦ (ks ∪ ks0) ∪ ks1 = ks ∪ ks0 ∪ ks1; from specification set-union[node-index]%unc: ⊦ ks ∪ ks0 = ks0 ∪ ks; from specification set-union[node-index]% ⊦ ¬ key ∈ ks → (key ∈ ks ∪ ks0 ↔ key ∈ ks0); (s)