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Proof step 6 from proof in-1dim-09 (previous step)

∃ sk. r -- sk ∈ st ⊦ ∃ sk. r -- sk ∈ addarray(r0, val, i, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
r -- sk ∈ st, r ∈ st ⊦ ∃ sk. r -- sk ∈ addarray(r0, val, i, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st])]]