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Proof step 9 from proof disj-write (previous step)

∀ r0. ¬ r0 ∈ H0 ∨ r0 ≠ r ∧ ¬ r0 ∈ H1 ⊦ r = null, r ∈ H1, (¬ r0 ∈ H0 ∨ ¬ r0 ∈ H1) ∧ ¬ r ∈ H0

User applied rule case distinction

And obtained the following new goals

Proof step 10:
r0 ∈ H0, r0 ∈ H1, ∀ r0. ¬ r0 ∈ H0 ∨ r0 ≠ r ∧ ¬ r0 ∈ H1 ⊦ r = null, r ∈ H1
Proof step 12:
r ∈ H0, ∀ r0. ¬ r0 ∈ H0 ∨ r0 ≠ r ∧ ¬ r0 ∈ H1 ⊦ r = null, r ∈ H1

Rule argument

pos: right 3

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