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Proof step 38 from proof precise-disj (conclusion)

precise(P0), precise(P1) ⊦ (P0 * P1 * true)(H0 ∪ H2) ∧ H0 ∪ H2 = H1 ∪ H3 ∧ H0 ⊥ H2 ∧ H1 ⊥ H3 ∧ P0(H0) ∧ P1(H1) → H0 ⊥ H1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
H0 ∪ H2 = H1 ∪ H3, (P0 * P1 * true)(H1 ∪ H3), H0 ⊥ H2, H1 ⊥ H3, P0(H0), P1(H1), precise(P0), precise(P1) ⊦ H0 ⊥ H1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]