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Proof step 23 from proof ole-last (previous step)

y = y0 + a2 ', x0 + a2 ' = a1 ' + y, a1 ∈ x0, a0 < a2, ordered≤(a1 ' + y), a1 < a2, ∀ a, b. ordered≤(x0 + a ') → (ordered≤(x0 + a ' + b ') ↔ ¬ b < a) ⊦ y = [], a1 = a2, b < a0, b < a1, a1 < a0, ordered≤((a0 ' + x0) + a2 ' + b '), b < a2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
a0 < a2, a1 < a2, ordered≤(a1 ' + y0 + a2 '), ∀ a, b. ordered≤((a1 ' + y0) + a ') → (ordered≤((a1 ' + y0) + a ' + b ') ↔ ¬ b < a) ⊦ a1 = a2, a1 < a0, b < a1, b < a0, b < a2, ordered≤(a1 ' + y0 + a2 ' + b ')

Rule argument

Additional information

Used simplifier rules: cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x);%app-01: ⊦ ordered≤(x + y) → ordered≤(y);%le-r: ⊦ ordered≤(a ' + b ' + x) ↔ ¬ b < a ∧ ordered≤(b + x);%c: ⊦ a + x = a ' + x; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%e-02: ⊦ x + a ' ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%app-06: ⊦ x + a ' = y + b ' ↔ x = y ∧ a = b; from specification list