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Proof step 19 from proof bytevalues-eq (conclusion)

⊦ (∀ val. val ∈ vals ∨ val ∈ vals0 → is_bytevalue(val)) → (bytevalues(vals) = bytevalues(vals0) ↔ vals = vals0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
bytevalues(vals) = bytevalues(vals0), ∀ val. val ∈ vals ∨ val ∈ vals0 → is_bytevalue(val) ⊦ vals = vals0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]