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Proof step 15 from proof hasSharing-different-disjoint (conclusion)

⊦ ¬ hasSharing(r, st) ∧ path∃(str, r, r1, st) ∧ path∃(str, r, r2, st) ∧ ¬ path∃(str, r1, r2, st) ∧ ¬ path∃(str, r2, r1, st) → disjoint(str, r1 ' + r2 ', st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
path∃(str, r, r2, st), path∃(str, r, r1, st) ⊦ hasSharing(r, st), path∃(str, r1, r2, st), path∃(str, r2, r1, st), disjoint(str, r1 ' + r2 ', st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]