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Proof step 13 from proof pathex-subset (conclusion)

⊦ validref(str, r, ty, st) ∧ path∃(str, r, r1, st) ∧ st ⊆ st0 → path∃(str, r, r1, st0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
st ⊆ st0, path∃(str, r, r1, st), validref(str, r, ty, st) ⊦ path∃(str, r, r1, st0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]