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Proof step 3 from proof insert-subset-old (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ks ⊆ ks0 ↔ (∀ key. key ∈ ks → key ∈ ks0)

Rule argument

spec: set-basic, inst: node-index name: Subset seq: ⊦ ks ⊆ ks0 ↔ (∀ key. key ∈ ks → key ∈ ks0) subst: [ks0, ks] → [ks0 ∪ ks1, ks ++ key] flags: <all paths>

Additional information

Used simplifier rules: ⊦ key ∈ ks ++ key0 ↔ key = key0 ∨ key ∈ ks; (s)% ⊦ key ∈ ks ∪ ks0 ↔ ¬(¬ key ∈ ks ∧ ¬ key ∈ ks0); (ss)%una: ⊦ (ks ∪ ks0) ∪ ks1 = ks ∪ ks0 ∪ ks1; from specification set-union[node-index]%unc: ⊦ ks ∪ ks0 = ks0 ∪ ks; from specification set-union[node-index]%sub-in: ⊦ key ∈ ks ∧ ks ⊆ ks0 → key ∈ ks0; (forward) from specification set-basic[node-index]%ne-01: ⊦ key ∈ ks → ks ≠ ∅; (forward) from specification set-basic[node-index] Used terms: ks0 ∪ ks1, ks ++ key