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Proof step 8 from proof isbytelist-all (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 4:
is_bytevalue(val) ∧ is_bytelist(vals0), is_bytelist(vals0) ↔ (∀ val. val ∈ vals0 → is_bytevalue(val)) ⊦ ∀ val0. val0 = val ∨ val0 ∈ vals0 → is_bytevalue(val0)
Proof step 7:
∀ val0. val0 = val ∨ val0 ∈ vals0 → is_bytevalue(val0), is_bytelist(vals0) ↔ (∀ val. val ∈ vals0 → is_bytevalue(val)) ⊦ is_bytevalue(val) ∧ is_bytelist(vals0)

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]