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Proof step 5 from proof eq-union-disj (
previous step
)
zh0 ∪ zh = zh1 ∪ zh, zh0 ⊥ zh, zh1 ⊥ zh ⊦ ∀ r. (r ∈ zh0 ↔ r ∈ zh1) ∧ (r ∈ zh0 → zh0[r] = zh1[r])
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 6
:
zh0 ∪ zh = zh1 ∪ zh, zh1 ⊥ zh, zh0 ⊥ zh ⊦ (r ∈ zh0 ↔ r ∈ zh1) ∧ (r ∈ zh0 → zh0[r] = zh1[r])
Rule argument
Additional information