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Proof step 21 from proof exlist (previous step)

⊦ ∃ val1, val2. flatten(val ++ charval(jchar)) = val1 ++ val2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
⊦ flatten(val ++ charval(jchar)) = flatten(val) ++ charval(jchar)
Proof step 20:
⊦ ∃ val0, val1. flatten(val) ++ charval(jchar) = val0 ++ val1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]