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Proof step 16 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), a ∈ ax0 -1l (a0 × d0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ a ∈ ax ↔ (∃ d. a × d ∈ ax)
Proof step 18:
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)

Rule argument

name: inassoc rotate: F subst: [ax, a] → [ax0 -1l (a0 × d0), a]

Additional information

Used terms: ax0 -1l (a0 × d0), a