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Proof step 5 from proof in-1init-05 (conclusion)

⊦ r ∈ addarray(r, vals, j, st) ↔ vals ≠ [] ∨ r ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ addarray(r, [], i, st) = st
Proof step 2:
⊦ r ∈ st → r ∈ addarray(r0, vals, j, st)
Proof step 4:
⊦ vals = [], r ∈ addarray(r, vals, j, st), r ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]