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Proof step 9 from proof sources-not-transitive (conclusion)

d1 = cdom(co0), d0 = cdom(co), cdom(co0) ⇒ cdom(co), cdom(co) ⇒ d, ¬ cdom(co0) ⇒ d, cdom(co0) ≠ cdom(co), cdom(co0) ≠ d ⊦ ¬ (d1 ∈ sources(co ' + co0 ', d0) ∧ d0 ∈ sources(co ' + co0 ', d) → d1 ∈ sources(co ' + co0 ', d))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
cdom(co0) ⇒ cdom(co), cdom(co) ⇒ d, cdom(co0) ∈ sources(co ' + co0 ', cdom(co)) ∧ cdom(co) ∈ sources(co ' + co0 ', d) → cdom(co0) ∈ sources(co ' + co0 ', d) ⊦ cdom(co0) = d, cdom(co0) = cdom(co), cdom(co0) ⇒ d

Rule argument

[emptyarg]

Additional information

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