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Proof step 3 from proof disj-r (
previous step
)
r ∈ H1, r ∈ H0, H0 ⊥ H1 ⊦
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 4
:
⊦ H0 ⊥ H1 ↔ (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1)
Proof step 5
:
r ∈ H0, r ∈ H1, ∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1 ⊦
Rule argument
spec: <toplevel> name: disj seq: ⊦ H0 ⊥ H1 ↔ (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1) subst: [H1, H0] → [H1, H0] flags: <all paths>
Additional information
Used terms: H1, H0