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Proof step 10 from proof eq-union-disj (previous step)

zh0 ⊥ zh, zh1 ⊥ zh, ¬ r ∈ zh → (r ∈ zh0 ↔ r ∈ zh1), r ∈ zh0 ∨ r ∈ zh → (r ∈ zh0 ⊃ zh0[r];zh[r]) = (zh1 ∪ zh)[r] ⊦ (r ∈ zh0 ↔ r ∈ zh1) ∧ (r ∈ zh0 → zh0[r] = zh1[r])

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ r ∈ zh ∧ zh ⊥ zh0 → (r ∈ zh0 ↔ false); (s)% ⊦ zh ⊥ zh0 → (r ∈ zh ∪ zh0 ↔ r ∈ zh ∨ r ∈ zh0); (s)%r ∈ zh ∪ zh0 ⊦ zh ⊥ zh0 → (zh ∪ zh0)[r] = (r ∈ zh ⊃ zh[r];zh0[r]); (s)% ⊦ r ∈ zh ∧ zh0 ⊥ zh → (r ∈ zh0 ↔ false); (s)