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Proof step 10 from proof exactarray-put-07 (conclusion)

⊦ rk .ref ≠ val .refval → (exactarray(val, ty, i, st[rk, val0]) ↔ exactarray(val, ty, i, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ rk .ref = val .refval, exactarray(val, ty, i, st[rk, val0]) ↔ exactarray(val, ty, i, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]