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Proof step 16 from proof uniquelen (previous step)

unique_al(ax0), unique_al(ay0 + (a × d)' + az), ∀ ay0. (∀ a. a ∈ ax0 ↔ a ∈ ay0) ∧ unique_al(ay0) → # ax0 = # ay0, ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az) ⊦ # ax0 = # ay0 + # az, a ∈ ax0, a ∈ ay0, a ∈ az

User applied rule all left

And obtained the following new goals

Proof step 17:
unique_al(ax0), unique_al(ay0 + (a × d)' + az), (∀ a. a ∈ ax0 ↔ a ∈ ay0 + az) ∧ unique_al(ay0 + az) → # ax0 = #(ay0 + az), ∀ ay0. (∀ a. a ∈ ax0 ↔ a ∈ ay0) ∧ unique_al(ay0) → # ax0 = # ay0, ∀ a0. a0 ≠ a → (a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az) ⊦ # ax0 = # ay0 + # az, a ∈ ax0, a ∈ ay0, a ∈ az

Rule argument

pos: left 3 terms: ay0 + az abort: F discard: F computed: F

Additional information

Used terms: ay0 + az