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Proof step 2 from proof fac-zero-02 (previous step)

fac(i) = 0 ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ ¬ fac(i) < 1
Proof step 4:
¬ fac(i) < 1, fac(i) = 0 ⊦

Rule argument

spec: int-fac inst: name: fac-pos seq: ⊦ ¬ fac(i) < 1 subst: [i] → [i] flags: current, []

Additional information

Used terms: i