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Proof step 7 from proof length-from (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
# 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-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%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list