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Proof step 55 from proof exlist (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 53:
⊦ flatten(val ++ return(val0, ty)) = flatten(val) ++ return(val0, ty)
Proof step 54:
∀ val. ∃ val3, val2. flatten(val ++ val1) = val3 ++ val2 ⊦ ∃ val0, val2. flatten(val) ++ return(val1, ty) = val0 ++ val2

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]