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Proof step 12 from proof subset-uni (previous step)

H0 ⊆ H1, H1 \ H0 ⊥ H0, r ∈ H1 ∨ r ∈ H0 ⊦ (r ∈ H1 ∧ ¬ r ∈ H0 ⊃ H1[r];H0[r]) = H1[r]

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 13:
H0 ⊆ H1, H1 \ H0 ⊥ H0, r ∈ H1 ∨ r ∈ H0, ¬ r ∈ H1 ∨ r ∈ H0 ⊦ H0[r] = H1[r]

Rule argument

pos: right 1

Additional information