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Proof step 22 from proof del-uni (previous step)

H0 ⊥ H1 ⊦ r0 ≠ r ∧ (r0 ∈ H0 ∨ r0 ∈ H1) ↔ r0 ∈ (H0 -- r) ∪ (H1 -- r)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1)
Proof step 24:
H0 ⊥ H1 ⊦ r0 ≠ r ∧ (r0 ∈ H0 ∨ r0 ∈ H1) ↔ r0 ∈ (H0 -- r) ∪ (H1 -- r), H0 -- r ⊥ H1 -- r
Proof step 27:
H0 ⊥ H1, H0 -- r ⊥ H1 -- r ⊦ r0 ≠ r ∧ (r0 ∈ H0 ∨ r0 ∈ H1) ↔ r0 ∈ H0 -- r ∨ r0 ∈ H1 -- r

Rule argument

spec: <toplevel> name: in-uni seq: ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1) subst: [H1, H0, r] → [H1 -- r, H0 -- r, r0] flags: <all paths>

Additional information

Used terms: H1 -- r, H0 -- r, r0