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Proof step 6 from proof isbytelist-all (previous step)

val = val ∨ val ∈ vals0 → is_bytevalue(val), ∀ val0. val0 ∈ vals0 → is_bytevalue(val0), is_bytelist(vals0) ↔ (∀ val. val ∈ vals0 → is_bytevalue(val)) ⊦ is_bytevalue(val) ∧ is_bytelist(vals0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
is_bytevalue(val), val0 ∈ vals0, ∀ val0. val0 ∈ vals0 → is_bytevalue(val0) ⊦ is_bytelist(vals0), is_bytevalue(val0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]