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Proof step 14 from proof uniquelen (previous step)

unique_al(ay), unique_al(ax0), a ∈ ay, ∀ ay0. unique_al(ay0) ∧ (∀ a. a ∈ ax0 ↔ a ∈ ay0) → # ax0 = # ay0, ∀ a0. a0 = a ∨ a0 ∈ ax0 ↔ a0 ∈ ay ⊦ 1 + # ax0 = # ay, a ∈ ax0

User applied rule insert lemma

And obtained the following new goals

Proof step 15:
⊦ a ∈ ax ↔ (∃ d, ay, az. ax = ay + (a × d)' + az)
Proof step 16:
a ∈ ay ↔ (∃ d, ay0, az. ay = ay0 + (a × d)' + az), unique_al(ax0), unique_al(ay), a ∈ ay, ∀ a0. a0 = a ∨ a0 ∈ ax0 ↔ a0 ∈ ay, ∀ ay0. unique_al(ay0) ∧ (∀ a. a ∈ ax0 ↔ a ∈ ay0) → # ax0 = # ay0 ⊦ 1 + # ax0 = # ay, a ∈ ax0

Rule argument

spec: <toplevel> name: ex seq: ⊦ a ∈ ax ↔ (∃ d, ay, az. ax = ay + (a × d)' + az) subst: [a, ax] → [a, ay]

Additional information

Used terms: a, ay