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Proof step 20 from proof boolval (previous step)

⊦ flatten(val ++ boolval(false)) = flatten(val) ++ boolval(false)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 19:
⊦ false, flatten(val ++ boolval(false)) = flatten(val) ++ boolval(false)

Rule argument

[fma-arg: false]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (false)]]