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Proof step 3 from proof orderedls-ls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
b < a0, ordered<(a0 ' + x0), a < a0, b < a, b ∈ x0, ∀ a1, b0. ¬ ordered<(a1 ' + x0) ∨ ¬ b0 < a1 ∨ ¬ b0 ∈ x0 ⊦ b = a0

Rule argument

Additional information

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