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Proof step 20 from proof to-to-leq (previous step)

n0 < # x, ∀ m, x. m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x → x to n0 to m = x to m ⊦ ((x to n0) + x[n0] ' to n0) + x[n0] ' = (x to n0) + x[n0] ', x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
n0 < # x, ∀ m, x. m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x → x to n0 to m = x to m ⊦ (x to n0) + x[n0] ' to n0 = x to n0, x = []

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = z + y ↔ x = z; (s)