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Proof step 46 from proof nextl-nextj-lemma-butlast (previous step)

⊦ ∃ lf, jj. lf(0) = a ' ∧ jj(0) = n0 ∧ lf(1) = nextl(jj(0), lf(0).last) ∧ jj(1) = nextj(jj(0), lf(0).last)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: λ n. (n = 0 ⊃ a ';nextl(n0, a)), λ n. (n = 0 ⊃ n0;nextj(n0, a)) abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ a ' = b ' ↔ a = b; (s)% ⊦ a '.last = a; (s) Used terms: λ n. (n = 0 ⊃ a ';nextl(n0, a)), λ n. (n = 0 ⊃ n0;nextj(n0, a))