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Proof step 50 from proof nextl-nextj-lemma-butlast (previous step)

lf(0) = a ', ∀ n. n ≤ m0 → lf(n + 1) = nextl(jj(n), lf(n).last) ∧ jj(n + 1) = nextj(jj(n), lf(n).last) ⊦ ∃ lf, jj0. lf(0) = a ' ∧ jj0(0) = jj(0) ∧ (∀ n. n ≤ m0 + 1 → lf(n + 1) = nextl(jj0(n), lf(n).last) ∧ jj0(n + 1) = nextj(jj0(n), lf(n).last))

User applied rule exists right

And obtained the following new goals

Proof step 51:
lf(0) = a ', ∀ n. n ≤ m0 → lf(n + 1) = nextl(jj(n), lf(n).last) ∧ jj(n + 1) = nextj(jj(n), lf(n).last) ⊦ (0 = m0 + 2 ⊃ nextl(jj(m0 + 1), lf(m0 + 1).last);lf(0)) = a ' ∧ (0 = m0 + 2 ⊃ nextj(jj(m0 + 1), lf(m0 + 1).last);jj(0)) = jj(0) ∧ (∀ n. n ≤ m0 + 1 → (n + 1 = m0 + 2 ⊃ nextl(jj(m0 + 1), lf(m0 + 1).last);lf(n + 1)) = nextl((n = m0 + 2 ⊃ nextj(jj(m0 + 1), lf(m0 + 1).last);jj(n)), (n = m0 + 2 ⊃ nextl(jj(m0 + 1), lf(m0 + 1).last);lf(n)).last) ∧ (n + 1 = m0 + 2 ⊃ nextj(jj(m0 + 1), lf(m0 + 1).last);jj(n + 1)) = nextj((n = m0 + 2 ⊃ nextj(jj(m0 + 1), lf(m0 + 1).last);jj(n)), (n = m0 + 2 ⊃ nextl(jj(m0 + 1), lf(m0 + 1).last);lf(n)).last))

Rule argument

pos: right 1 terms: λ n. (n = m0 + 2 ⊃ nextl(jj(m0 + 1), lf(m0 + 1).last);lf(n)), λ n. (n = m0 + 2 ⊃ nextj(jj(m0 + 1), lf(m0 + 1).last);jj(n)) abort: F discard: T computed: F

Additional information

Used terms: λ n. (n = m0 + 2 ⊃ nextl(jj(m0 + 1), lf(m0 + 1).last);lf(n)), λ n. (n = m0 + 2 ⊃ nextj(jj(m0 + 1), lf(m0 + 1).last);jj(n))