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Proof step 8 from proof precise-same (conclusion)

precise(P) ⊦ P(H0) ∧ P(H1) ∧ 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 7:
H0 ∪ H = H1 ∪ H2, P(H0), P(H1), H0 ⊥ H, H1 ⊥ H2, precise(P), H0 ≠ H1 ∨ H ≠ H2 ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]