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Proof step 1 from proof subset-put-strong-02 (conclusion)

⊦ st ⊆ st0 → st[a, d][a0, d0][a1, d1] ⊆ st0[a, d][a0, d0][a1, d1]

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: subset-put-strong-05: ⊦ ¬ st[a, d][a0, d0][a1, d1] ⊆ st0[a, d][a0, d0][a1, d1] → ¬ st ⊆ st0;