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Proof step 2 from proof okarray-subset-02 (conclusion)

⊦ okarray(r, ty, st) ∧ ¬ okarray(r, ty, st0) → ¬ st ⊆ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ okarray(r, ty, st) ∧ st ⊆ st0 → okarray(r, ty, st0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]