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Proof step 8 from proof precise-ls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
(H4 ∪ H3) ∪ H0 = H1 ∪ H, ((r => a × r1 * ls(r1, x0)))(H1), ((r => a × r0))(H4), ls(r0, x0)(H3), H1 ⊥ H, H4 ⊥ H3, H0 ⊥ H4, H0 ⊥ 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 = H1

Rule argument

Additional information

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