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Proof step 1 from proof sources-not-transitive (
previous step
)
⊦ d0 ∈ sources(co ' + co0 ', d) ↔ d0 = d ∨ d0 = cdom(co0) ∧ d0 ⇒ d ∨ d0 = cdom(co) ∧ d0 ⇒ d ∨ d0 = cdom(co) ∧ d0 ⇒ cdom(co0) ∧ cdom(co0) ⇒ d
Lemma
sources-in-two