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Proof step 14 from proof precise-sep (previous step)

(H5 ∪ H1) ∪ H0 = (H4 ∪ H3) ∪ H, P0(H5), P0(H4), P1(H1), P1(H3), precise(P0), precise(P1), H0 ⊥ H1, H0 ⊥ H5, H4 ⊥ H3, H1 ⊥ H5, H ⊥ H3, H ⊥ H4 ⊦ H5 = H4 ∧ H1 = H3

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
(H5 ∪ H1) ∪ H0 = (H4 ∪ H3) ∪ H, P0(H4), P0(H5), P1(H3), P1(H1), precise(P1), precise(P0), H ⊥ H4, H ⊥ H3, H1 ⊥ H5, H4 ⊥ H3, H0 ⊥ H5, H0 ⊥ H1 ⊦ H5 = H4 ∧ H1 = H3

Proof step 14 from proof precise-sep (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information