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Proof step 5 from proof intval (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ val. flatten(val ++ intval(i0)) = flatten(val) ++ intval(i0) ⊦ flatten(val ++ intval(i0 + 1)) = flatten(val) ++ intval(i0 + 1)

Rule argument

[emptyarg]

Additional information

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