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Proof step 6 from proof append-from-gt (previous step)

∀ y, n0. x0 + y from n0 + # x0 = y from n0 ∨ n0 = 0 ⊦ (a + x0) + y from n0 + #(a + x0) = y from n0, n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ y, n0. x0 + y from n0 + # x0 = y from n0 ∨ n0 = 0 ⊦ x0 + y from n0 + # x0 = y from n0, n0 = 0

Rule argument

Additional information

Used simplifier rules: from-succ-cons-l-lem: ⊦ a ' + x from n + 1 = x from n;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)