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Proof step 5 from proof lenequal (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 6:
# ax0 + 1 = # ay, unique_al(ay), b ∈ ay, unique_al(ax0), ∀ ay0, b0. unique_al(ay0) ∧ # ax0 = # ay0 ∧ (∀ a. a ∈ ax0 → a ∈ ay0) ∧ b0 ∈ ay0 → b0 ∈ ax0, a = a ∨ a ∈ ax0 → a ∈ ay, ∀ a0. a0 ∈ ax0 → a0 ∈ ay ⊦ b = a, b ∈ ax0, a ∈ ax0

Rule argument

pos: left 6 terms: a abort: F discard: F computed: T

Additional information

Used terms: a