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Proof step 18 from proof min-com (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 19:
x3 + a0 ' + x = [] ∨ min(x3 + a0 ' + x) = min((a0 ' + x) + x3), min(x + a ' + x3) = min((a ' + x3) + x), a0 < min((a ' + x3) + x), a < min(x3 + a0 ' + x), ∀ x1, x2. #(x1 + x2) < #((a + x3) + (a0 + x)) → true → x1 + x2 = [] ∨ min(x1 + x2) = min(x2 + x1) ⊦ a = a0

Rule argument

subst: [x1, x2] → [x3, a0 ' + x]

Additional information

Used simplifier rules: len: ⊦ #(x + y) = # x + # y; 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%c: ⊦ a + x = a ' + x; 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