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Proof step 5 from proof ocs-del-length (previous step)

a = a0, ∀ a, b. #oc(a, x0 -l b) = #oc(a, x0) ∨ a = b ⊦ a = b, #oc(a, (a0 ' + x0) -l b) = #oc(a, a0 ') + #oc(a, x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ a, b. #oc(a, x0 -l b) = #oc(a, x0) ∨ a = b ⊦ a0 = b, #oc(a0, x0 -l b) = #oc(a0, x0)

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%one: ⊦ #oc(a, a ') = 1;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%del-n: ⊦ a ≠ b → (b ' + x) -l a = b ' + (x -l a); from specification list-del