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Proof step 3 from proof owns-one-map (previous step)

r ∈ rs, ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦ # rs = 0, r = null, rs = ∅, ∃ o. (g(r)[r => o] * true)[rs, f, g]

User applied rule exists right

And obtained the following new goals

Proof step 4:
r ∈ rs, ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦ # rs = 0, r = null, rs = ∅, (g(r)[r => f(r)] * true)[rs, f, g]

Rule argument

pos: right 4 terms: f(r) abort: F discard: T computed: F

Additional information

Used terms: f(r)