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Proof step 13 from proof to-to-leq (previous step)

n0 < #((x to n0) + x[n0] '), n0 < # x, ∀ m, x. m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x → x to n0 to m = x to m ⊦ x = [], ((x to n0) + x[n0] ' to n0) + ((x to n0) + x[n0] ')[n0] ' = x to n0 + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
n0 < # x, ∀ m, x. m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x → x to n0 to m = x to m ⊦ x = [], ((x to n0) + x[n0] ' to n0) + ((x to n0) + x[n0] ')[n0] ' = x to n0 + 1

Rule argument

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%len-02: ⊦ # a ' = 0 +1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list