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Proof step 5 from proof ocs-del-del1 (previous step)

#oc(a, a0 ') + #oc(a, x0) = 1, ∀ a. #oc(a, x0) = 1 → x0 -l a = x0 -1l a ⊦ (a0 ' + x0) -l a = (a0 ' + x0) -1l a, a = a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
#oc(a, x0) = 1, ∀ a. #oc(a, x0) = 1 → x0 -l a = x0 -1l a ⊦ x0 -l a = x0 -1l a, a = a0

Rule argument

Additional information

Used simplifier rules: one-01: ⊦ a ≠ b → #oc(a, b ') = 0;% ⊦ x + y = x + z ↔ y = z; (s)% ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a); (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ≠ b → (b ' + x) -l a = b ' + (x -l a); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)