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Proof step 17 from proof precise-disj (previous step)

H ∪ H4 = H2 ∪ H0, ((P1 * true))(H4), P0(H), P1(H2), precise(P0), precise(P1), H2 ⊥ H0, H ⊥ H4, ∃ H2, H1. H2 ⊥ H1 ∧ H0 = H2 ∪ H1 ∧ P0(H2) ∧ true(H1) ⊦ H ⊥ H2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
H ∪ H4 = H2 ∪ H3 ∪ H1, ((P1 * true))(H4), P0(H), P0(H3), P1(H2), precise(P1), precise(P0), H3 ⊥ H1, H ⊥ H4, H2 ⊥ H3, H2 ⊥ H1 ⊦ H ⊥ H2

Rule argument

Additional information

Used simplifier rules: trueH: ⊦ true(H);% ⊦ H1 ⊥ H → (H0 ⊥ H1 ∪ H ↔ H0 ⊥ H1 ∧ H0 ⊥ H); (ws)