つぎは,定理シェマである:
(0) | num=num1 ⊃ num+num11=num1+num11
| (1) | num+num1′= num′+num1
| (2) | num+num1 = num1+num
| (3) | (num+num1)+num11 = num+(num1+num11)
| (4) | num×num1 = num1×num
| (5) | (num×num1)×num11 = num×(num1×num11)
| (6) | (num+num1)×num11 = num×num11+num1×num11
|
証明の例として,以下,(2) の証明を示す。
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先ず,1+1=1+1 は定理:
| | ε
| | | ↓(E1)
| 1+1=1+1 | ←─ | trm=trm
|
-
演繹定理を用いて
x+1=1+x → x′+1=1+x′
を証明する:
x+1=1+x | ←─ | num=num1
| | | ↓(N0)
| (x+1)′=(1+x)′ | ←─ | num′=num1′
| ‖
| (x+1)′=(1+x)′ | ←─ | (num+1)′=trm
| | | ↓(A1)
| (x′)′=(1+x)′ | ←─ | (num′)′=trm
| ‖
| (x′)′=(1+x)′ | ←─ | num′=trm
| | | ↓(A1)
| x′+1=(1+x)′ | ←─ | num+1=trm
| ‖
| x′+1=(1+x)′ | ←─ | trm=(num+num1)′
| | | ↓(A2)
| x′+1=1+x′ | ←─ | trm=num+num1′
|
-
x+1=1+x の証明:
1+1=1+1 | ←─ | sen[nvr↓1]
| & | | &
| x+1=1+n⊃x′+1=1+x′ | | sen⊃sen[nvr↓nvr′]
| | | ↓(N3)
| x+1=1+x | ←─ | sen
|
-
演繹定理を用いて
x+b=b+x ⊃ x+b′=b′+x
を証明する。
x+b=b+x | ←─ | num=num1
| | | ↓(N0)
| (x+b)′=(b+x)′ | ←─ | num′=num1′
| ‖
| (x+b)′=(b+x)′ | ←─ | (num+num1)′=trm
| | | ↓(A2)
| x+b′=(b+x)′ | ←─ | num+num1′=trm
| ‖
| x+b′=(b+x)′ | ←─ | trm=(num+num1)′
| | | ↓(A2)
| x+b′=b+x′ | ←─ | trm=num+num1′
| | | ↓(1)
| x+b′=b+x′ | ←─ | trm=trm1
| & | | &
| b+x′=b′+x | | trm1=trm11
| | | ↓(E3)
| x+b′=b′+x | ←─ | trm=trm11
|
-
最後に,a+b=b+a がつぎのように証明される:
a+1=1+a | ←─ | sen[nvr↓1]
| & | | &
| a+x=x+a⊃a+x′=x′+a | | sen⊃sen[nvr↓nvr′]
| | | ↓(N3)
| a+x=x+a | ←─ | sen
| | | ↓(Q0)
| a+b=b+a | ←─ | sen[tvr↓trm]
|
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