Step | Hyp | Ref
| Expression |
1 | | ianor 474 |
. . . 4
⊢ (¬ (A = 0c ∧ B =
0c) ↔ (¬ A =
0c ∨ ¬ B = 0c)) |
2 | | nc0suc 6218 |
. . . . . . . 8
⊢ (A ∈ NC → (A =
0c ∨ ∃p ∈ NC A = (p
+c 1c))) |
3 | 2 | ord 366 |
. . . . . . 7
⊢ (A ∈ NC → (¬ A =
0c → ∃p ∈ NC A = (p +c
1c))) |
4 | 3 | adantr 451 |
. . . . . 6
⊢ ((A ∈ NC ∧ B ∈ NC ) → (¬ A
= 0c → ∃p ∈ NC A = (p +c
1c))) |
5 | | addc32 4417 |
. . . . . . . . 9
⊢ ((p +c 1c)
+c B) = ((p +c B) +c
1c) |
6 | | 0cnsuc 4402 |
. . . . . . . . 9
⊢ ((p +c B) +c 1c) ≠
0c |
7 | 5, 6 | eqnetri 2534 |
. . . . . . . 8
⊢ ((p +c 1c)
+c B) ≠
0c |
8 | | addceq1 4384 |
. . . . . . . . . 10
⊢ (A = (p
+c 1c) → (A +c B) = ((p
+c 1c) +c B)) |
9 | 8 | eqeq1d 2361 |
. . . . . . . . 9
⊢ (A = (p
+c 1c) → ((A +c B) = 0c ↔ ((p +c 1c)
+c B) =
0c)) |
10 | 9 | necon3bbid 2551 |
. . . . . . . 8
⊢ (A = (p
+c 1c) → (¬ (A +c B) = 0c ↔ ((p +c 1c)
+c B) ≠
0c)) |
11 | 7, 10 | mpbiri 224 |
. . . . . . 7
⊢ (A = (p
+c 1c) → ¬ (A +c B) = 0c) |
12 | 11 | rexlimivw 2735 |
. . . . . 6
⊢ (∃p ∈ NC A = (p
+c 1c) → ¬ (A +c B) = 0c) |
13 | 4, 12 | syl6 29 |
. . . . 5
⊢ ((A ∈ NC ∧ B ∈ NC ) → (¬ A
= 0c → ¬ (A
+c B) =
0c)) |
14 | | nc0suc 6218 |
. . . . . . . 8
⊢ (B ∈ NC → (B =
0c ∨ ∃p ∈ NC B = (p
+c 1c))) |
15 | 14 | ord 366 |
. . . . . . 7
⊢ (B ∈ NC → (¬ B =
0c → ∃p ∈ NC B = (p +c
1c))) |
16 | 15 | adantl 452 |
. . . . . 6
⊢ ((A ∈ NC ∧ B ∈ NC ) → (¬ B
= 0c → ∃p ∈ NC B = (p +c
1c))) |
17 | | addcass 4416 |
. . . . . . . . 9
⊢ ((A +c p) +c 1c) =
(A +c (p +c
1c)) |
18 | | 0cnsuc 4402 |
. . . . . . . . 9
⊢ ((A +c p) +c 1c) ≠
0c |
19 | 17, 18 | eqnetrri 2536 |
. . . . . . . 8
⊢ (A +c (p +c 1c)) ≠
0c |
20 | | addceq2 4385 |
. . . . . . . . . 10
⊢ (B = (p
+c 1c) → (A +c B) = (A
+c (p
+c 1c))) |
21 | 20 | eqeq1d 2361 |
. . . . . . . . 9
⊢ (B = (p
+c 1c) → ((A +c B) = 0c ↔ (A +c (p +c 1c)) =
0c)) |
22 | 21 | necon3bbid 2551 |
. . . . . . . 8
⊢ (B = (p
+c 1c) → (¬ (A +c B) = 0c ↔ (A +c (p +c 1c)) ≠
0c)) |
23 | 19, 22 | mpbiri 224 |
. . . . . . 7
⊢ (B = (p
+c 1c) → ¬ (A +c B) = 0c) |
24 | 23 | rexlimivw 2735 |
. . . . . 6
⊢ (∃p ∈ NC B = (p
+c 1c) → ¬ (A +c B) = 0c) |
25 | 16, 24 | syl6 29 |
. . . . 5
⊢ ((A ∈ NC ∧ B ∈ NC ) → (¬ B
= 0c → ¬ (A
+c B) =
0c)) |
26 | 13, 25 | jaod 369 |
. . . 4
⊢ ((A ∈ NC ∧ B ∈ NC ) → ((¬ A = 0c
∨ ¬ B = 0c)
→ ¬ (A +c
B) =
0c)) |
27 | 1, 26 | syl5bi 208 |
. . 3
⊢ ((A ∈ NC ∧ B ∈ NC ) → (¬ (A = 0c ∧ B =
0c) → ¬ (A
+c B) =
0c)) |
28 | 27 | con4d 97 |
. 2
⊢ ((A ∈ NC ∧ B ∈ NC ) → ((A
+c B) =
0c → (A =
0c ∧ B = 0c))) |
29 | | addceq12 4386 |
. . 3
⊢ ((A = 0c ∧ B =
0c) → (A
+c B) =
(0c +c
0c)) |
30 | | addcid2 4408 |
. . 3
⊢
(0c +c 0c) =
0c |
31 | 29, 30 | syl6eq 2401 |
. 2
⊢ ((A = 0c ∧ B =
0c) → (A
+c B) =
0c) |
32 | 28, 31 | impbid1 194 |
1
⊢ ((A ∈ NC ∧ B ∈ NC ) → ((A
+c B) =
0c ↔ (A =
0c ∧ B = 0c))) |