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Proof step 11 from proof precise-ls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
(H4 ∪ H3) ∪ H0 = (H5 ∪ H2) ∪ H, ((r => a × r0))(H4), ((r => a × r1))(H5), ls(r0, x0)(H3), ls(r1, x0)(H2), H4 ⊥ H3, H0 ⊥ H4, 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 ⊦ H4 ∪ H3 = H5 ∪ H2

Rule argument

Additional information

Used simplifier rules: ⊦ H1 ⊥ H2 → (H1 ∪ H2 ⊥ H0 ↔ H0 ⊥ H1 ∧ H0 ⊥ H2); (ws)