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Proof step 4 from proof rmtoobigpos (previous step)

#(a + x0) ≤ n, ∀ n. # x0 ≤ n → x0 -p n = x0 ⊦ (a + x0) -p n = a + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# x0 < n, ∀ n. # x0 ≤ n → x0 -p n = x0 ⊦ (a ' + x0) -p n = a ' + x0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (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)