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Proof step 2 from proof perm-same-prefix (previous step)

perm(y, z), perm(z, y) ⊦ ∃ x, y0, z0. y = x + y0 ∧ z = x + z0 ∧ perm(y0, z0)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], y, z abort: F discard: T computed: F

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: [], y, z