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Proof step 44 from proof introduce-sharing (
previous step
)
P(H), P0(H1 ∪ H), P1(H2 ∪ H), H2 ⊥ H, H1 ⊥ H2, H1 ⊥ H ⊦ ∃ H0, H3, H4. H0 ⊥ H4 ∧ H0 ⊥ H3 ∧ H3 ⊥ H4 ∧ H1 ∪ H2 ∪ H = H0 ∪ H3 ∪ H4 ∧ P0(H0 ∪ H4) ∧ P1(H3 ∪ H4)
User applied rule exists right
And closed the goal!
Rule argument
pos: right 1 terms: H1, H2, H abort: F discard: T computed: T
Additional information
Used terms: H1, H2, H