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Proof step 4 from proof ad (previous step)

a ∈ y, #oc(a0, (x + y) -1l a) < #oc(a0, x) + #oc(a0, y -1l a) ⊦ #oc(a0, x) + #oc(a0, y -1l a) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: del1-01 seq: ⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x) subst: [b, x, a] → [a, x + y, a0] flags: rewrite, current

Additional information

Used simplifier rules: del1: ⊦ a ∈ x → #oc(a, x -1l a) = #oc(a, x) -1;%oc-del-01: ⊦ a ∈ y → #oc(a, (x + y) -1l a) = #oc(a, x + y) -1;%ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%del1-01: ⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x);%ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%a: ⊦ (x + y) + z = x + y + z; from specification list%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%pred-03: ⊦ n -1 = n - 1; from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2 Used terms: a, x + y, a0