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Proof step 8 from proof length-from (previous step)

#(a + x0) = n0 + n1 + 1, ∀ x, n0. # x = n0 + n1 → #(x from n1) = n0, ∀ n0, n1. # x0 = n0 + n1 + 1 ∧ (∀ x, n0. # x = n0 + n1 → #(x from n1) = n0) → #(x0 from n1 + 1) = n0 ⊦ #(a + x0 from n1 + 1) = n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# x0 = n0 + n1, ∀ n0, n1. # x0 = n0 + n1 + 1 ∧ (∀ x, n0. # x = n0 + n1 → #(x from n1) = n0) → #(x0 from n1 + 1) = n0, ∀ x, n0. # x = n0 + n1 → #(x from n1) = n0 ⊦ #(x0 from n1) = n0

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% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)