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Proof step 4 from proof intvalues-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ intvalues(val ' + vals) = val .intval + intvalues(vals)
Proof step 3:
∀ vals0. intvalues(vals1 + vals0) = intvalues(vals1) + intvalues(vals0) ⊦ intvalues(vals1 + vals0) = intvalues(vals1) + intvalues(vals0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ ints + ints0 = ints + ints1 ↔ ints0 = ints1], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ i + ints = i ' + ints], [CSIMPSEQ : ⊦ val + vals = val ' + vals])]]