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Proof step 18 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, a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az, a ∈ ax0, a ∈ ay0, a ∈ az

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 19:
a0 ∈ ax0, 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, a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az, a ∈ ax0, a ∈ ay0, a ∈ az
Proof step 21:
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) ⊦ a0 ∈ ax0, # ax0 = # ay0 + # az, a0 ∈ ax0 ↔ a0 ∈ ay0 ∨ a0 ∈ az, a ∈ ax0, a ∈ ay0, a ∈ az

Rule argument

fma: a0 ∈ ax0

Additional information

Used formulas: a0 ∈ ax0