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Proof step 2 from proof rest (previous step)

⊦ (∀ a0. #oc(a0, a ' + x) ≤ #oc(a0, b ' + x)) ↔ a = b

User applied rule case distinction

And obtained the following new goals

Proof step 3:
∀ a0. #oc(a0, a ' + x) ≤ #oc(a0, b ' + x) ⊦ a = b

Rule argument

pos: right 1

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic