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Proof step 8 from proof intval (previous step)

i0 = i +1, ∀ val. flatten(val ++ intval(i0)) = flatten(val) ++ intval(i0) ⊦ flatten(val ++ intval(i)) = flatten(val) ++ intval(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ val0. flatten(val0 ++ intval(i + 1)) = flatten(val0) ++ intval(i + 1) ⊦ flatten(val ++ intval(i)) = flatten(val) ++ intval(i)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i +1 = i + 1])]]