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Proof step 40 from proof exlist (previous step)

∀ val. ∃ val3, val2. flatten(val ++ val1) = val3 ++ val2, ∀ val. ∃ val1, val3. flatten(val ++ val2) = val1 ++ val3 ⊦ ∃ val4, val3. flatten(val ++ (val1 ++ val2)) = val4 ++ val3

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
⊦ flatten(val ++ (val0 ++ val1)) = flatten(val ++ val0 ++ val1)
Proof step 39:
∀ val. ∃ val3, val2. flatten(val ++ val1) = val3 ++ val2, ∀ val. ∃ val1, val3. flatten(val ++ val2) = val1 ++ val3 ⊦ ∃ val0, val3. flatten(val ++ val1 ++ val2) = val0 ++ val3

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]