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Proof step 17 from proof disjoint-pathex-forward (conclusion)

⊦ disjoint(str, r ' + r0 ', st) ∧ path∃(str, r, r1, st) ∧ r ≠ r0 → disjoint(str, r1 ' + r0 ', st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
path∃(str, r, r1, st), disjoint(str, r ' + r0 ', st) ⊦ r = r0, disjoint(str, r1 ' + r0 ', st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]