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Proof step 8 from proof precise-split-l (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: H0, H2, H1, H abort: F discard: F computed: T

Additional information

Used simplifier rules: ⊦ H ⊥ H0 ∧ H ⊥ H1 → (H ∪ H0 = H ∪ H1 ↔ H0 = H1); (s) Used terms: H0, H2, H1, H