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Proof step 21 from proof pathex-subset-equiv (conclusion)

⊦ ¬ junkref(str, r, st) ∧ st ⊆ st0 → (path∃(str, r, r0, st0) ↔ path∃(str, r, r0, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
st ⊆ st0 ⊦ path∃(str, r, r0, st0) ↔ path∃(str, r, r0, st), junkref(str, r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]