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Proof step 7 from proof bytevalues-eq (previous step)

bytevalues(vals0) = bytevalues(vals), (val ∈ vals0 ∨ val ∈ vals → is_bytevalue(val)) → vals0 = vals, val1 ∈ vals0 ∨ val1 = val1 ∨ val1 ∈ vals → is_bytevalue(val1), ∀ val0. val0 ∈ vals0 ∨ val0 ∈ vals → is_bytevalue(val0) ⊦ byteval(val1 .byteval) ' + vals0 = val1 ' + vals

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
bytevalues(vals0) = bytevalues(vals), is_bytevalue(val1), ∀ val0. val0 ∈ vals0 ∨ val0 ∈ vals → is_bytevalue(val0), val ∈ vals0 ∨ val ∈ vals ⊦ vals0 = vals, is_bytevalue(val)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ vals + vals0 = vals + vals1 ↔ vals0 = vals1], [CSIMPSEQ : is_bytevalue(val) ⊦ byteval(val .byteval) = val])]]