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Proof step 14 from proof mk-exists (previous step)

∀ H1. # H1 < # (mk(rs, f) ++ r)[r, o] → true → (∃ rs, f. H1 = mk(rs, f)) ⊦ r = null, r ∈ rs, ∃ rs0, f0. (mk(rs, f) ++ r)[r, o] = mk(rs0, f0)

User applied rule exists right

And obtained the following new goals

Proof step 15:
∀ H1. # H1 < # (mk(rs, f) ++ r)[r, o] → true → (∃ rs, f. H1 = mk(rs, f)) ⊦ (mk(rs, f) ++ r)[r, o] = mk(rs ++ r, f(r; o)), r = null, r ∈ rs

Rule argument

pos: right 3 terms: rs ++ r, f(r; o) abort: F discard: T computed: F

Additional information

Used terms: rs ++ r, f(r; o)