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Proof step 2 from proof look-out (conclusion)

⊦ (st[r -- _type, val][_out]) = st[_out]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ (st[rk0, val][rk]) = ((rk = rk0 ⊃ val; st[rk]))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- _type ≠ _out])]]