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Proof step 12 from proof precise-disj (previous step)

H1 ∪ H3 = H ∪ H4, ((P1 * true))(H4), P0(H), P1(H1), precise(P1), precise(P0), H ⊥ H4, H1 ⊥ H3, ∃ H0, H1. H0 ⊥ H1 ∧ H ∪ H4 = H0 ∪ H1 ∧ P1(H0) ∧ ((true * P0))(H1) ⊦ H ⊥ H1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
H ∪ H4 = H2 ∪ H0, ((true * P0))(H0), ((P1 * true))(H4), P0(H), P1(H2), precise(P0), precise(P1), H2 ⊥ H0, H ⊥ H4 ⊦ H ⊥ H2

Rule argument

Additional information

Used simplifier rules: precise-same: precise(P) ⊦ P(H0) ∧ P(H1) ∧ H0 ⊥ H ∧ H1 ⊥ H2 → (H0 ∪ H = H1 ∪ H2 ↔ H0 = H1 ∧ H = H2);