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Proof step 3 from proof pathex-subset-01 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
st ⊆ st0, validref(str, r, ty, st), path∃(str, r, r1, st) ⊦ path∃(str, r, r1, st0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]