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Proof step 2 from proof leninv-less (previous step)

leninv(nσ) < n∞ ⊦ ∃ n. leninv(nσ) = ⌜n⌝ ∧ ⌜n⌝ < n∞

User applied rule exists right

And obtained the following new goals

Proof step 3:
leninv(nσ) < n∞ ⊦ leninv(nσ) = ⌜leninv(nσ).n⌝ ∧ ⌜leninv(nσ).n⌝ < n∞, ∃ n. leninv(nσ) = ⌜n⌝ ∧ ⌜n⌝ < n∞

Rule argument

pos: right 1 terms: leninv(nσ).n abort: F discard: F computed: F

Additional information

Used terms: leninv(nσ).n