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Proof step 3 from proof subset-uni (previous step)

H0 ⊆ H1 ⊦ ∃ H. H ∪ H0 = H1 ∧ H ⊥ H0

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: H1 \ H0 abort: F discard: T computed: F

Additional information

Used terms: H1 \ H0