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Proof step 1 from proof disj-alloc (conclusion)
r0 ≠ null, ¬ r0 ∈ H0, ¬ r0 ∈ H1 ⊦ H0 ⊥ H1 → H0 ⊥ H1 ++ r0
System applied rule normalize
And obtained the following new goals
Proof step 2
:
H0 ⊥ H1 ⊦ r0 = null, r0 ∈ H0, r0 ∈ H1, H0 ⊥ H1 ++ r0
Rule argument
Additional information