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

ordered≤(x) ⊦ a = b, ins≤(a, x -1l b) = ins≤(a, x) -1l b

User applied rule structural induction

And obtained the following new goals

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

Rule argument

indvar: x

Additional information

Used simplifier rules: ins-e: ⊦ ins≤(a, []) = a ';%list freely generated by [], +;;% ⊦ [] -1l a = []; (s)% ⊦ a ≠ b → a ' -1l b = a '; (s)% ⊦ a ' = b ' ↔ a = b; (s)