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Proof step 11 from proof uniquelen (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 12:
unique_al(ay), unique_al(ax0), ∀ ay0. unique_al(ay0) ∧ (∀ a. a ∈ ax0 ↔ a ∈ ay0) → # ax0 = # ay0, a ∈ ax0 ∨ a = a ↔ a ∈ ay, ∀ a0. a0 ∈ ax0 ∨ a0 = a ↔ a0 ∈ ay ⊦ 1 + # ax0 = # ay, a ∈ ax0

Rule argument

pos: left 4 terms: a abort: F discard: F computed: T

Additional information

Used terms: a