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Proof step 2 from proof ocs-same-length-mklist (previous step)

∀ a. #oc(a, x0) = # x0 ↔ x0 = mklist(a, # x0) ⊦ #oc(a, a0 + x0) = #(a0 + x0) ↔ a0 + x0 = mklist(a, #(a0 + x0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a. #oc(a, x0) = # x0 ↔ x0 = mklist(a, # x0) ⊦ #oc(a, a0 ') + #oc(a, x0) = # x0 + 1 ↔ a0 ' + x0 = mklist(a, # x0 + 1)

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list