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Proof step 40 from proof nextl-lemma-butlast (previous step)

lf(0) = a ', ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ⊦ ∃ lf. lf(0) = a ' ∧ (∀ m. ¬ n0 < m → lf(m + 1) = nextl(m, lf(m).last))

User applied rule exists right

And obtained the following new goals

Proof step 41:
lf(0) = a ', ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ⊦ (0 = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(0)) = a ' ∧ (∀ m. ¬ n0 < m → (m + 1 = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m + 1)) = nextl(m, (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)).last))

Rule argument

pos: right 1 terms: λ n. (n = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(n)) abort: F discard: T computed: F

Additional information

Used terms: λ n. (n = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(n))