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Proof step 2 from proof precise-same (previous step)

P(H0), P(H1), precise(P), H0 ⊥ H, H1 ⊥ H2 ⊦ H0 ∪ H = H1 ∪ H2 ↔ H0 = H1 ∧ H = H2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
H0 ∪ H = H1 ∪ H2, P(H1), P(H0), precise(P), H1 ⊥ H2, H0 ⊥ H ⊦ H0 = H1 ∧ H = H2

Rule argument

Additional information