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Proof step 5 from proof length-from (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# x = n0 + n1 + 1, ∀ x, n0. # x = n0 + n1 → #(x from n1) = n0 ⊦ #(x from n1 + 1) = n0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2