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Proof step 4 from proof isbytelist-all (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
is_bytevalue(val), is_bytelist(vals0), val0 ∈ vals0, ∀ val. val ∈ vals0 → is_bytevalue(val) ⊦ is_bytevalue(val0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]