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Proof step 11 from proof del1-01-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
ordered≤(a0 ' + x0), ∀ a1, b0. a1 = b0 ∨ ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ⊦ a = b, ins≤(a, (a0 ' + x0) -1l b) = a ' + ((a0 ' + x0) -1l b), a0 < a

Rule argument

Additional information

Used simplifier rules: ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x;% ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a); (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)