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Proof step 2 from proof stsubset (conclusion)

⊦ newref(r, st) → st ⊆ addarray(r, ty, i, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ newref(r, st) → (st ⊆ addarray(r, ty, i, st0) ↔ st ⊆ st0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ st ⊆ st])]]