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Proof step 11 from proof bytevalues-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
by ' + bytevalues(vals0) = bytevalues(vals1), by ∈ bytevalues(vals1), ∀ val0. val0 ∈ vals0 ∨ val0 ∈ vals1 → is_bytevalue(val0), bytevalues(vals0) = bytevalues(vals1 .rest) ∧ (val ∈ vals0 ∨ val ∈ vals1 .rest → is_bytevalue(val)) → vals0 = vals1 .rest ⊦ byteval(by) ' + vals0 = vals1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]