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Proof step 5 from proof to-plus-from-inf-simp (previous step)

⊦ (n0 < n ⊃ nf(n0);nf((n0 - n) + n)) = nf(n0)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 6:
¬ n0 < n ⊦ nf((n0 - n) + n) = nf(n0)

Rule argument

pos: right 1

Additional information