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Proof step 16 from proof ordered-postfix (previous step)

ordered≤(a0 ' + y), a1 ' + x1 = a ' + a0 ' + y ∨ a1 ' + x1 < a ' + a0 ' + y, ∀ x0. x0 ⊒ a0 ' + y → ordered≤(x0) ⊦ a0 < a, ordered≤(a1 ' + x1)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 2

Additional information

Used simplifier rules: le-r: ⊦ ordered≤(a ' + b ' + x) ↔ ¬ b < a ∧ ordered≤(b + x);%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification oelem%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%c: ⊦ a + x = a ' + x; from specification list