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Proof step 13 from proof subset-uni (previous step)

H0 ⊆ H1, H1 \ H0 ⊥ H0, r ∈ H1 ∨ r ∈ H0, ¬ r ∈ H1 ∨ r ∈ H0 ⊦ H0[r] = H1[r]

User applied rule case distinction

And obtained the following new goals

Proof step 14:
r ∈ H0, H0 ⊆ H1, H1 \ H0 ⊥ H0, r ∈ H1 ∨ r ∈ H0 ⊦ H0[r] = H1[r]

Rule argument

pos: left 4

Additional information

Used simplifier rules: subset-in: ⊦ H0 ⊆ H1 ∧ r ∈ H0 → r ∈ H1;