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Proof step 24 from proof max-com (previous step)

max(x + a ' + x3) = (max(x3 + x) < a ⊃ a;max(x3 + x)), max(x3 + a0 ' + x) = (max(x + x3) < a0 ⊃ a0;max(x + x3)), ∀ x1, x2. # x1 + # x2 < #(a + x3) + #(a0 + x) → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ x + x3 = [], x3 + x = [], ((max(x + x3) < a0 ⊃ a0;max(x + x3)) < a ⊃ a;(max(x + x3) < a0 ⊃ a0;max(x + x3))) = ((max(x3 + x) < a ⊃ a;max(x3 + x)) < a0 ⊃ a0;(max(x3 + x) < a ⊃ a;max(x3 + x)))

User applied rule apply induction

And obtained the following new goals

Proof step 25:
max(x3 + a0 ' + x) = (max(x + x3) < a0 ⊃ a0;max(x + x3)), max(x + a ' + x3) = (max(x3 + x) < a ⊃ a;max(x3 + x)), x + x3 = [] ∨ max(x + x3) = max(x3 + x), ∀ x1, x2. # x1 + # x2 < #(a + x3) + #(a0 + x) → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ x3 + x = [], x + x3 = [], ((max(x + x3) < a0 ⊃ a0;max(x + x3)) < a ⊃ a;(max(x + x3) < a0 ⊃ a0;max(x + x3))) = ((max(x3 + x) < a ⊃ a;max(x3 + x)) < a0 ⊃ a0;(max(x3 + x) < a ⊃ a;max(x3 + x)))

Rule argument

subst: [x1, x2] → [x, x3]

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s)% ⊦ #(a ' + x) = (# x)+1; (s)% ⊦ m +1 = m + 1; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s) Used terms: x, x3