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Proof step 2 from proof newref-subset-01 (conclusion)

⊦ newref(r, st1) ∧ st ⊆ st1 → newref(r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ ¬ newref(r, st) ∧ newref(r, st0) → ¬ st ⊆ st0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]