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Proof step 51 from proof completeness (conclusion)
⊦ 〈build#(x; to)〉 (to = fail → ¬ (∃ t. x = depths(t, 0)))
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 50
:
⊦ 〈build#(x; to)〉 (to = fail → (∀ t. x ≠ depths(t, 0)))
Rule argument
[emptyarg]
Additional information
[[CONCRETESIMPRULES : ()]]