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Proof step 3 from proof ordered-postfix (previous step)

∀ x. x ⊒ x1 ∧ ordered≤(x1) → ordered≤(x), x ⊒ a + x1, ordered≤(a + x1) ⊦ ordered≤(x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
ordered≤(a ' + x1), x ⊒ a ' + x1, ∀ x0. x0 ⊒ x1 → ordered≤(x0) ⊦ ordered≤(x)

Rule argument

Additional information

Used simplifier rules: app-01: ⊦ ordered≤(x + y) → ordered≤(y);%cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x);%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list