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Proof step 22 from proof precise-sep (previous step)

(H5 ∪ H1) ∪ H0 = (H4 ∪ H3) ∪ H, P0(H4), P0(H5), P1(H3), P1(H1), H4 ⊥ H, H ⊥ H3, H1 ⊥ H5, H4 ⊥ H3, H0 ⊥ H5, H0 ⊥ H1, ∀ H, H0, H1, H2. P0(H) ∧ P0(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0, ∀ H, H0, H1, H2. P1(H) ∧ P1(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0 ⊦ H5 = H4 ∧ H1 = H3

User applied rule all left

And obtained the following new goals

Proof step 23:
(H5 ∪ H1) ∪ H0 = (H4 ∪ H3) ∪ H, P0(H5), P0(H4), P1(H1), P1(H3), H0 ⊥ H1, H0 ⊥ H5, H4 ⊥ H3, H1 ⊥ H5, H ⊥ H3, H4 ⊥ H, P0(H5) ∧ P0(H4) ∧ H5 ⊥ H1 ∪ H0 ∧ H4 ⊥ H3 ∪ H ∧ H5 ∪ H1 ∪ H0 = H4 ∪ H3 ∪ H → H5 = H4, ∀ H, H0, H1, H2. P1(H) ∧ P1(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0 ⊦ H5 = H4 ∧ H1 = H3

Rule argument

pos: left 12 terms: H5, H4, H1 ∪ H0, H3 ∪ H abort: F discard: T computed: F

Additional information

Used terms: H5, H4, H1 ∪ H0, H3 ∪ H