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Proof step 7 from proof precise-same (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: H0, H1, H, H2 abort: F discard: T computed: T

Additional information

Used terms: H0, H1, H, H2