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Proof step 13 from proof precise-ls (previous step)

H4 = H5, r0 = r1, (H4 ∪ H3) ∪ H0 = (H5 ∪ H2) ∪ H, ((r => a × r1))(H5), ((r => a × r0))(H4), ls(r1, x0)(H2), ls(r0, x0)(H3), H ⊥ H2, H ⊥ H5, H5 ⊥ H2, H0 ⊥ H3, H0 ⊥ H4, H4 ⊥ H3, ∀ H2, H0, H1, H, r. H2 ∪ H0 = H1 ∪ H ∧ ls(r, x0)(H2) ∧ ls(r, x0)(H1) ∧ H2 ⊥ H0 ∧ H1 ⊥ H → H2 = H1 ⊦ H4 ∪ H3 = H5 ∪ H2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
(H5 ∪ H3) ∪ H0 = (H5 ∪ H2) ∪ H, ((r => a × r1))(H5), ls(r1, x0)(H3), ls(r1, x0)(H2), H5 ⊥ H3, H0 ⊥ H5, H0 ⊥ H3, H5 ⊥ H2, H ⊥ H5, H ⊥ H2, ∀ H2, H0, H1, H, r. H2 ∪ H0 = H1 ∪ H ∧ ls(r, x0)(H2) ∧ ls(r, x0)(H1) ∧ H2 ⊥ H0 ∧ H1 ⊥ H → H2 = H1 ⊦ H3 = H2

Rule argument

Additional information

Used simplifier rules: precise(P) ⊦ P(H0) ∧ P(H1) ∧ H0 ⊥ H ∧ H1 ⊥ H2 → (H0 ∪ H = H1 ∪ H2 ↔ H0 = H1 ∧ H = H2); (s)% ⊦ precise(r => ce); (s)