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Proof step 48 from proof nextl-nextj-lemma-butlast (previous step)

∀ a, n0, nextl, nextj. ∃ lf, jj. lf(0) = a ' ∧ jj(0) = n0 ∧ (∀ n. n ≤ m0 → lf(n + 1) = nextl(jj(n), lf(n).last) ∧ jj(n + 1) = nextj(jj(n), lf(n).last)) ⊦ ∃ lf, jj. lf(0) = a ' ∧ jj(0) = n0 ∧ (∀ n. n ≤ m0 + 1 → lf(n + 1) = nextl(jj(n), lf(n).last) ∧ jj(n + 1) = nextj(jj(n), lf(n).last))

User applied rule all left

And obtained the following new goals

Proof step 49:
∃ lf, jj. lf(0) = a ' ∧ jj(0) = n0 ∧ (∀ n. n ≤ m0 → lf(n + 1) = nextl(jj(n), lf(n).last) ∧ jj(n + 1) = nextj(jj(n), lf(n).last)) ⊦ ∃ lf, jj. lf(0) = a ' ∧ jj(0) = n0 ∧ (∀ n. n ≤ m0 + 1 → lf(n + 1) = nextl(jj(n), lf(n).last) ∧ jj(n + 1) = nextj(jj(n), lf(n).last))

Rule argument

pos: left 1 terms: a, n0, nextl, nextj abort: F discard: T computed: F

Additional information

Used terms: a, n0, nextl, nextj