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Proof step 4 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 cut applied rule cut formula

And obtained the following new goals

Proof step 3:
⊦ ¬ is_valuelist(val0) → flatten(val ++ val0) = flatten(val) ++ val0

Rule argument

[fma-arg: is_valuelist(intval(i0 + 1))]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ is_valuelist(intval(i))], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ val ++ val1 = val0 ++ val1 ↔ val = val0])], [EXTRAFMAS : (is_valuelist(intval(i0 + 1)))]]