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Proof step 6 from proof uni-ass (previous step)

H ⊥ H0, H1 ⊥ H, H0 ⊥ H1 ⊦ (r ∈ (H0 ∪ H1) ∪ H ↔ r ∈ H0 ∪ H1 ∪ H) ∧ (r ∈ (H0 ∪ H1) ∪ H → ((H0 ∪ H1) ∪ H)[r] = (H0 ∪ H1 ∪ H)[r])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1)
Proof step 8:
H0 ⊥ H1, H1 ⊥ H, H ⊥ H0, H0 ∪ H1 ⊥ H ⊦ (r ∈ H0 ∪ H1 ∨ r ∈ H ↔ r ∈ H0 ∪ H1 ∪ H) ∧ (r ∈ H0 ∪ H1 ∨ r ∈ H → ((H0 ∪ H1) ∪ H)[r] = (H0 ∪ H1 ∪ H)[r])

Rule argument

spec: <toplevel> name: in-uni seq: ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1) subst: [H1, H0, r] → [H, H0 ∪ H1, r] flags: <all paths>

Additional information

Used simplifier rules: disj-distr-uni-com: ⊦ H1 ⊥ H2 → (H1 ∪ H2 ⊥ H0 ↔ H0 ⊥ H1 ∧ H0 ⊥ H2);%disj-com-rw: ⊦ H0 ⊥ H1 → H1 ⊥ H0; Used terms: H, H0 ∪ H1, r