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Proof step 2 from proof bytevalues (previous step)

∀ vals1. bytevalues(vals0 + vals1) = bytevalues(vals0) + bytevalues(vals1) ⊦ bytevalues((val + vals0) + vals1) = bytevalues(val + vals0) + bytevalues(vals1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
∀ vals1. bytevalues(vals0 + vals1) = bytevalues(vals0) + bytevalues(vals1) ⊦ bytevalues(vals0 + vals1) = bytevalues(vals0) + bytevalues(vals1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1], [CSIMPSEQ : ⊦ bytes + bytes0 = bytes + bytes1 ↔ bytes0 = bytes1], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ by + bytes = by ' + bytes], [CSIMPSEQ : ⊦ bytevalues(val ' + vals) = val .byteval + bytevalues(vals)], [CSIMPSEQ : ⊦ val + vals = val ' + vals])]]