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Proof step 9 from proof isbytelist-all (previous step)

is_bytelist(vals0) ↔ (∀ val. val ∈ vals0 → is_bytevalue(val)) ⊦ is_bytelist(val + vals0) ↔ (∀ val0. val0 ∈ val + vals0 → is_bytevalue(val0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ is_bytelist(val ' + vals) ↔ is_bytevalue(val) ∧ is_bytelist(vals)
Proof step 8:
is_bytelist(vals0) ↔ (∀ val. val ∈ vals0 → is_bytevalue(val)) ⊦ is_bytevalue(val) ∧ is_bytelist(vals0) ↔ (∀ val0. val0 = val ∨ val0 ∈ vals0 → is_bytevalue(val0))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ val ∈ val0 ' ↔ val = val0], [CSIMPSEQ : ⊦ val ∈ vals + vals0 ↔ ¬ ¬ (val ∈ vals ∨ val ∈ vals0)], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ val + vals = val ' + vals])]]