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Proof step 20 from proof addarray-both-subset (conclusion)

⊦ ¬ addarray(r, byte_type, vals, st0) ⊆ addarray(r, byte_type, vals, st1) → ¬ st0 ⊆ st1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
st0 ⊆ st1 ⊦ addarray(r, byte_type, vals, st0) ⊆ addarray(r, byte_type, vals, st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]