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Proof step 3 from proof precise-same (previous step)

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

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

Proof step 4:
H0 ∪ H = H1 ∪ H2, P(H0), P(H1), precise(P), H0 ⊥ H, H1 ⊥ H2 ⊦ H0 = H1, H0 = H1 ∧ H = H2

Rule argument

fma: H0 = H1

Additional information

Used simplifier rules: ⊦ H ⊥ H0 ∧ H ⊥ H1 → (H ∪ H0 = H ∪ H1 ↔ H0 = H1); (s) Used formulas: H0 = H1