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Proof step 8 from proof lenequal (previous step)

1 + # ax0 = # ay, unique_al(ay), unique_al(ax0), b ∈ ay, a ∈ ay, ∀ ay0, b0. # ax0 = # ay0 ∧ unique_al(ay0) ∧ b0 ∈ ay0 ∧ (∀ a. a ∈ ax0 → a ∈ ay0) → b0 ∈ ax0, ∀ a0. a0 ∈ ax0 → a0 ∈ ay ⊦ b = a, b ∈ ax0, a ∈ ax0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ a ∈ ax ↔ (∃ d, ay, az. ax = ay + (a × d)' + az)
Proof step 10:
1 + # ax0 = # ay, unique_al(ax0), unique_al(ay), b ∈ ay, ∀ a0. a0 ∈ ax0 → a0 ∈ ay, ∀ ay0, b0. # ax0 = # ay0 ∧ unique_al(ay0) ∧ b0 ∈ ay0 ∧ (∀ a. a ∈ ax0 → a ∈ ay0) → b0 ∈ ax0, ∃ d, ay0, az. ay = ay0 + (a × d)' + az ⊦ b = a, a ∈ ax0, b ∈ ax0

Rule argument

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

Additional information

Used terms: ay, a