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

ordered≤(a ' + (x -1l a)), ordered≤(x), a ∈ x ⊦ a ' + (x -1l a) = x

User applied rule structural induction

And closed the goal!

Rule argument

indvar: x

Additional information

Used simplifier rules: app-01: ⊦ ordered≤(x + y) → ordered≤(y);%cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x);%le-r: ⊦ ordered≤(a ' + b ' + x) ↔ ¬ b < a ∧ ordered≤(b + x);%ole-in: ⊦ ordered≤(a ' + x) ∧ b ∈ x → ¬ b < a;%list freely generated by [], +;;% ⊦ a + x = a ' + x; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ∈ b ' ↔ a = b; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x) -1l a = x; (s)% ⊦ x + y = x + z ↔ y = z; (s)% ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a); (s)% ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss)% ⊦ [] -1l a = []; (s)% ⊦ x + [] = x; (s)% ⊦ a ' ≠ []; (s)% ⊦ ¬ a ∈ []; (s)