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Proof step 28 from proof precise-ls (previous step)

H3 ∪ H0 = H2 ∪ H, ((r => a × r1))(H5), ls(r1, x0)(H2), ls(r1, x0)(H3), H ⊥ H2, H0 ⊥ H3, H5 ⊥ H3, H0 ⊥ H5, H5 ⊥ H2, H ⊥ H5, H5 ⊥ H3 ∪ H0, H5 ⊥ H2 ∪ H, ∀ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
H3 ∪ H0 = H2 ∪ H, ((r => a × r1))(H5), ls(r1, x0)(H3), ls(r1, x0)(H2), H5 ⊥ H2 ∪ H, H ⊥ H5, H5 ⊥ H2, H0 ⊥ H5, H5 ⊥ H3, H0 ⊥ H3, 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

Proof step 28 from proof precise-ls (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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