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Proof step 3 from proof del1-01-02 (previous step)

∀ a, b. ordered≤(x0) → ins≤(a, x0 -1l b) = ins≤(a, x0) -1l b ∨ a = b, ordered≤(a0 + x0) ⊦ ins≤(a, (a0 + x0) -1l b) = ins≤(a, a0 + x0) -1l b, a = b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
ordered≤(a0 ' + x0), ∀ a1, b0. ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ∨ a1 = b0 ⊦ ins≤(a, (a0 ' + x0) -1l b) = ins≤(a, a0 ' + x0) -1l b, a = b

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