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Proof step 2 from proof del1-01-01 (previous step)

a ∈ x, ordered≤(x), ordered≤(a ' + (x -1l a)) ⊦ 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);%del1-01: ⊦ ordered≤(x) → ordered≤(x -1l a);%le-r: ⊦ ordered≤(a ' + b ' + x) ↔ ¬ b < a ∧ ordered≤(b + x);%orderedle-ls: ⊦ b < a ∧ b ∈ x → ¬ ordered≤(a ' + x);%list freely generated by [], +;;%c: ⊦ a + x = a ' + x; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%del1-y: ⊦ (a ' + x) -1l a = x; from specification list-del%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%del1-n: ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a); from specification list-del%sls: ⊦ a ≠ b → (¬ a < b ↔ b < a); from specification oelem%del1-e: ⊦ [] -1l a = []; from specification list-del%app: ⊦ x + [] = x; from specification list%e: ⊦ a ' ≠ []; from specification list%in-empty: ⊦ ¬ a ∈ []; from specification list