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Proof step 8 from proof bytevalues-eq (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 7:
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

Rule argument

[EXRARG : pos: left 3, [extquanttermlist: (val1)abort:F, discard:F, computed:T]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val1)]]