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Proof step 2 from proof ole-01 (previous step)

∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦ ordered≤(ins≤(a, a0 + x0)) ↔ ordered≤(a0 + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦ ordered≤(ins≤(a, a0 ' + x0)) ↔ ordered≤(a0 ' + x0)

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x; from specification list