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

b < a, ordered<(a ' + (a0 + x0)), b ∈ a0 + x0, ∀ b, a. ¬(b < a ∧ ordered<(a ' + x0) ∧ b ∈ x0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ls-r: ⊦ ordered<(a ' + b ' + x) ↔ a < b ∧ ordered<(b + x);% ⊦ ¬ a < a; (s)%transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a + x = a ' + x; (s)