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Proof step 4 from proof orderedls-ls (previous step)

b < a0, ordered<(a0 ' + x0), a < a0, b < a, b ∈ x0, ∀ a1, b0. ¬ ordered<(a1 ' + x0) ∨ ¬ b0 < a1 ∨ ¬ b0 ∈ x0 ⊦ b = a0

Heuristic quantifier closing applied rule all left

And obtained the following new goals

Proof step 5:
b < a0, ordered<(a0 ' + x0), a < a0, b < a, b ∈ x0, ¬ ordered<(b ' + x0) ∨ ¬ a0 < b ∨ ¬ a0 ∈ x0, ∀ a1, b0. ¬ ordered<(a1 ' + x0) ∨ ¬ b0 < a1 ∨ ¬ b0 ∈ x0 ⊦ b = a0

Rule argument

pos: left 6 terms: b, a0

Additional information

Used terms: b, a0