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Proof step 12 from proof bytevalues-eq (previous step)

by ' + bytevalues(vals0) = bytevalues(vals1), by ∈ bytevalues(vals1), ∀ val0. val0 ∈ vals0 ∨ val0 ∈ vals1 → is_bytevalue(val0), ∀ vals1. bytevalues(vals0) = bytevalues(vals1) ∧ (∀ val. val ∈ vals0 ∨ val ∈ vals1 → is_bytevalue(val)) → vals0 = vals1 ⊦ byteval(by) ' + vals0 = vals1

User applied rule all left

And obtained the following new goals

Proof step 11:
by ' + bytevalues(vals0) = bytevalues(vals1), by ∈ bytevalues(vals1), ∀ val0. val0 ∈ vals0 ∨ val0 ∈ vals1 → is_bytevalue(val0), bytevalues(vals0) = bytevalues(vals1 .rest) ∧ (∀ val. val ∈ vals0 ∨ val ∈ vals1 .rest → is_bytevalue(val)) → vals0 = vals1 .rest ⊦ byteval(by) ' + vals0 = vals1

Rule argument

[EXRARG : pos: left 4, [extquanttermlist: (vals1 .rest) abort:F , discard:T , computed:F ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (vals1 .rest)]]