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Proof step 5 from proof subset-put-notin (previous step)

⊦ st ⊆ st0 ∧ ¬ r -- sk ∈ st ∨ (r -- sk) × val ∈ st ∧ st -- (r -- sk) ⊆ st0 -- (r -- sk) ↔ st ⊆ st0, r ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk × val ∈ st ↔ rk ∈ st ∧ st[rk] = val])]]