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Proof step 18 from proof unique-del (previous step)

a × d1 ∈ ax0, unique_al(ax0), unique_al(ax0 -1l (a0 × d0)) ⊦ a0 = a, a0 × d0 = a × d, a0 ∈ ax0 -1l (a0 × d0), ∃ d. a × d ∈ ax0 -1l (a0 × d0)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: d1 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ ¬ p ∈ ax -1l p0 ∧ p ∈ ax → p = p0; (forward)% ⊦ a × d = a0 × d0 ↔ a = a0 ∧ d = d0; Used terms: d1