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Proof step 11 from proof max-com (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 12:
x3 + (a0 + x) = [] ∨ max(x3 + (a0 + x)) = max((a0 + x) + x3), ∀ x1, x2. # x1 + # x2 < #(a + x3) + #(a0 + x) → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ (max(x3 + a0 ' + x) < a ⊃ a;max(x3 + a0 ' + x)) = max((a0 ' + x) + a ' + x3)

Rule argument

subst: [x1, x2] → [x3, a0 + x]

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x; from specification list%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2 Used terms: x3, a0 + x