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Proof step 4 from proof path-nodups-subset (previous step)

¬ (path(x0, H) ∧ ¬ r ∈ x0), r ∈ H, path(r ' + x0, H), # H ≤ # x0 ⊦ dups(x0), r ∈ x0, path(x0, H -- r)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
r ∈ H, path(r ' + x0, H), # H ≤ # x0 ⊦ dups(x0), r ∈ x0, path(x0, H -- r), path(x0, H)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]